Q1 )
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\)th of the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter.

(iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound Interest at 8% per annum.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

(i)Taxi fare for 1st km = Rs 15,

Taxi fare after 2 km = 15 + 8 = Rs 23

Taxi fare after 3 km = 23 + 8 = Rs 31

Taxi fare after 4 km = 31 + 8 = Rs 39

Therefore, the sequence is 15, 23, 31, 39…

It is an arithmetic progression because difference between any two consecutive terms is equal which is 8.

(23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, …)

(ii)Let amount of air initially present in a cylinder = V

Amount of air left after pumping out air by vacuum pump

= \( V - {{V} \over {4}} \)

\( = {{4V - V} \over {4}} \)

\( = {{3V} \over {4}}\)

Amount of air left when vacuum pump again pumps out air

=\({{3V} \over {4}} - ( {{1} \over {4}} × {{3V} \over {4}} ) \)

\( = {{3V} \over {4}} - {{3V} \over {16}} \)

\( = {{12V - 3V} \over {16}} \)

\( = {{9V} \over {16}}\)

So, the sequence we get is like \(V, {{3V} \over {4}}, {{9V} \over {16}}.....\)

Checking for difference between consecutive terms …

\({{3V} \over {4}} - V = {{-V} \over {4}}\),

\({{9V} \over {16}} - {{3V} \over {4}} = {{9V - 12V} \over {16}} = {{-3V} \over {16}}\)

Difference between consecutive terms is not equal.

Therefore, it is not an arithmetic progression.

(iii) Cost of digging 1 meter of well = Rs 150

Cost of digging 2 meters of well = 150 + 50 = Rs 200

Cost of digging 3 meters of well = 200 + 50 = Rs 250

Therefore, we get a sequence of the form 150, 200, 250…

It is an arithmetic progression because difference between any two consecutive terms is equal. (200 – 150 = 250 – 200 = 50…)

Here, difference between any two consecutive terms (common difference) is equal to 50.

(iv)Amount in bank after Ist year = \(1000 ( 1 + {{8} \over {100}})\)… (1)

Amount in bank after two years = \(1000 ( 1 + {{8} \over {100}})^2\)… (2)

Amount in bank after three years = \(1000 ( 1 + {{8} \over {100}})^3\)… (3)

Amount in bank after four years = \(1000 ( 1 + {{8} \over {100}})^4\)… (4)

It is not an arithmetic progression because \((2) - (1) \ne (3) - (2)\)

(Difference between consecutive terms is not equal)

Therefore, it is not an Arithmetic Progression.

Q2 )
Write first four terms of the AP, when the first term a and common difference d are given as follows:

(i)a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = -3

(iv) a = -1, d = \({{1} \over {2}}\)

(v) a = -1.25, d = -0.25

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

(i) First term = a = 10, d = 10

Second term = a + d = 10 + 10 = 20

Third term = second term + d = 20 + 10 = 30

Fourth term = third term + d = 30 + 10 = 40

Therefore, first four terms are: 10, 20, 30, 40

(ii) First term = a = –2 , d = 0

Second term = a + d = –2 + 0 = –2

Third term = second term + d = –2 + 0 = –2

Fourth term = third term + d = –2 + 0 = –2

Therefore, first four terms are: –2, –2, –2, –2

(iii) First term = a = 4, d =–3

Second term = a + d = 4 – 3 = 1

Third term = second term + d = 1 – 3 = –2

Fourth term = third term + d = –2 – 3 = –5

Therefore, first four terms are: 4, 1, –2, –5

(iv) First term = a = –1, d = \({{1} \over {2}}\)

Second term = a + d = –1 + \({{1} \over {2}}\) = \({{-1} \over {2}}\)

Third term = second term + d = \({{-1} \over {2}} + {{1} \over {2}}\) = 0

Fourth term = third term + d = 0 + \({{1} \over {2}}\) = \({{1} \over {2}}\)

Therefore, first four terms are: –1, \({{-1} \over {2}}\), 0, \({{1} \over {2}}\)

(v) First term = a = –1.25, d = –0.25

Second term = a + d = –1.25 – 0.25 = –1.50

Third term = second term + d = –1.50 – 0.25 = –1.75

Fourth term = third term + d = –1.75 – 0.25 = –2.00

Therefore, first four terms are: –1.25, –1.50, –1.75, –2.00

Q3 )
For the following APs, write the first term and the common difference.

(i) 3, 1, –1, –3 …

(ii) –5, –1, 3, 7…

(iii) \({{1}\over{3}},{{5}\over{3}},{{9}\over{3}},{{13}\over{3}}\)

(iv) 0.6, 1.7, 2.8, 3.9 …

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

(i) 3, 1, –1, –3…

First term = a = 3,

Common difference (d)

= Second term – first term

= Third term – second term

and so on

Therefore, Common difference (d) = 1 – 3 = –2

(ii) –5, –1, 3, 7…

First term = a = –5

Common difference (d)

= Second term – First term

= Third term – Second term

and so on

Therefore, Common difference (d)

= –1 – (–5) = –1 + 5 = 4

(iii)\({{1}\over{3}},{{5}\over{3}},{{9}\over{3}},{{13}\over{3}}\)

First term = a = \({{1}\over{3}}\)

Common difference (d)

= Second term – First term

= Third term – Second term

and so on

Therefore, Common difference (d)

= \({{5}\over{3}} - {{1}\over{3}} = {{4}\over{3}}\)

(iv) 0.6, 1.7, 2.8, 3.9…

First term = a = 0.6

Common difference (d)

= Second term – First term

= Third term – Second term and so on

Therefore, Common difference (d)

= 1.7 - 0.6 = 1.1

Q4 )
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

(i) \(2, 4, 8, 16…\)

(ii) 2, \({{5}\over{2}}\) , 3, \({{7}\over{2}}\)…

(iii) \(-1.2, -3.2, -5.2, -7.2…\)

(iv) \(-10, -6, -2, 2…\)

(v) \(3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}\)…

(vi) \(0.2, 0.22, 0.222, 0.2222…\)

(vii) \(0, -4, -8, -12…\)

(viii) \({{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}\)…

(ix) \(1, 3, 9, 27…\)

(x)\( a, 2a, 3a, 4a…\)

(xi) \(a , a^2 , a^3 , a^4 \)…

(xii) \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32} \) …

(xiii) \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12} \)…

(xiv) \(1^2, 3^2, 5^2, 7^2\)…

(xv) \(1^2, 5^2, 7^2, 73\)…

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

(i) 2, 4, 8, 16…

It is not an AP because difference between consecutive terms is not equal.

4 – 2 \( \ne \) 8 - 4

(ii) 2, \({{5}\over{2}}\) , 3, \({{7}\over{2}}\)…

It is an AP because difference between consecutive terms is equal.

\(\Rightarrow {{5}\over{2}} - 2 = 3 - {{5}\over{2}} = {{1}\over{2}}\)

Common difference (d) = \({{1}\over{2}}\)

Fifth term = \({{7}\over{2}} + {{1}\over{2}} = 4\)

Sixth term = 4 + \({{1}\over{2}} = {{9}\over{2}}\)

Seventh term = \({{9}\over{2}} + {{1}\over{2}} = 5\)

Therefore, next three terms are 4, \({{9}\over{2}}\) and 5.

(iii)-1.2, -3.2, -5.2, -7.2…

It is an AP because difference between consecutive terms is equal.

-3.2 - (-1.2)

= -5.2 - (-3.2)

= -7.2 - (-5.2) = -2

Common difference (d) = -2

Fifth term = -7.2 – 2 = -9.2

Sixth term = -9.2 – 2 = -11.2

Seventh term = -11.2 – 2 = -13.2

Therefore, next three terms are -9.2, -11.2 and -13.2

(iv) -10, -6, -2, 2…

It is an AP because difference between consecutive terms is equal.

-6 - (-10) = -2 - (-6)

= 2 - (-2) = 4

Common difference (d) = 4

Fifth term = 2 + 4 = 6

Sixth term = 6 + 4 = 10

Seventh term = 10 + 4 = 14

Therefore, next three terms are 6, 10 and 14

(v) 3 ,\(3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}\)…

It is an AP because difference between consecutive terms is equal.

= \(3 + \sqrt{2} - 3 \)

= 3 + 2\sqrt{2} - 3 + \sqrt{2}\)

\( = \sqrt{2}\)

Common difference (d) = \(\sqrt{2}\)

Fifth term

= \(3 + 3\sqrt{2} + \sqrt{2} \)

\( = 3 + 4\sqrt{2}\)

Sixth term

= \(3 + 4\sqrt{2} + \sqrt{2} \)

\( = 3 + 5\sqrt{2}\)

Seventh term

= \(3 + 5\sqrt{2} + \sqrt{2}\)

\( = 3 + 6\sqrt{2}\)

Therefore, next three terms are \( 3 + 4\sqrt{2}, 3 + 5\sqrt{2}. 3 + 6\sqrt{2}\)

(vi) 0.2, 0.22, 0.222, 0.2222…

It is not an AP because difference between consecutive terms is not equal.

0.22 - 0.2 \( \ne \) 0.222 - 0.22

(vii) 0, -4, -8, -12…

It is an AP because difference between consecutive terms is equal.

-4 – 0 = -8 - (-4)

= -12 - (-8) = -4

Common difference (d) = -4

Fifth term = -12 – 4 =-16

Sixth term = -16 – 4 = -20

Seventh term = -20 – 4 = -24

Therefore, next three terms are -16, -20 and -24

(viii) \({{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}\)…

It is an AP because difference between consecutive terms is equal.

=\({{-1}\over{2}} - {{-1}\over{2}} = {{-1}\over{2}} - {{-1}\over{2}} = 0\)

Common difference (d) = 0

Fifth term = \({{-1}\over{2}} + 0 = {{-1}\over{2}}\)

Sixth term = \({{-1}\over{2}} + 0 = {{-1}\over{2}}\)

Seventh term =\({{-1}\over{2}} + 0 = {{-1}\over{2}}\)

Therefore, next three terms are \({{-1}\over{2}},{{-1}\over{2}},{{-1}\over{2}}\)

(ix) 1, 3, 9, 27…

It is not an AP because difference between consecutive terms is not equal.

3 – 1 \( \ne \) 9 - 3

(x) a, 2a, 3a, 4a…

It is an AP because difference between consecutive terms is equal.

2a – a

= 3a - 2a

= 4a - 3a = a

Common difference (d) = a

Fifth term = 4a + a = 5a

Sixth term = 5a + a = 6a

Seventh term = 6a + a = 7a

Therefore, next three terms are 5a, 6a and 7a

(xi) \(a , a^2 , a^3 , a^4 \)…

It is not an AP because difference between consecutive terms is not equal

\(a^2 - a \ne a^3 - a^2\)

(xii) \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32} \)…

It is an AP because difference between consecutive terms is equal.

=\(\sqrt{2},2\sqrt{2},3\sqrt{2},4\sqrt{2}\)

Common difference (d) = \(\sqrt{2}\)

Fifth term = \(4\sqrt{2} + \sqrt{2} = 5\sqrt{2}\)

Sixth term = \(5\sqrt{2} + \sqrt{2} = 6\sqrt{2}\)

Seventh term =
\(6\sqrt{2} + \sqrt{2} = 7\sqrt{2}\)

Therefore, next three terms are \(5\sqrt{2}, 6\sqrt{2}, 7\sqrt{2}\)

(xiii) \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12} \)…

It is not an AP because difference between consecutive terms is not equal.

= \(\sqrt{6} - \sqrt{3} \ne \sqrt{9} - \sqrt{6}\)

(xiv) \(1^2, 3^2, 5^2, 7^2\)…

It is not an AP because difference between consecutive terms is not equal.

= \(3^2 - 1^2 \ne 5^2 - 3^2\)

(xv) \(1^2, 5^2, 7^2, 73\)…

1, 25, 49, 73…

It is an AP because difference between consecutive terms is equal.

= \( 5^2 - 1^2 \)

\(= 7^2 - 5^2\)

\( = 73 - 7^2\)

\( = 24\)

Common difference (d) = 24

Fifth term = 73 + 24 = 97

Sixth term = 97 + 24 = 121

Seventh term = 121 + 24 = 145

Therefore, next three terms are 97, 121 and 145

Q1 )
Find the missing variable from \( a, d, n\) and \(a_n\), where a is the first term, d is the common difference and \(a_n\) is the nth term of AP.

(i) \(a = 7, d = 3, n = 8\)

(ii) \(a = –18, n = 10, a_n = 0\)

(iii) \(d = –3, n = 18, a_n = -5\)

(iv) \(a = –18.9, d = 2.5, a_n = 3.6\)

(v) \(a = 3.5, d = 0, n = 105\)

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

(i) a = 7, d = 3, n = 8

We need to find \(a_n\) here.

Using formula \(a_n = a + (n - 1)d\)

Putting values of a, d and n,

\( a_n\) = 7 + (8 - 1) 3

= 7 + (7) 3 = 7 + 21 = 28

(ii) a = –18, n= 10, \(a_n = 0\)

We need to find d here.

Using formula \(a_n = a + (n - 1)d\)

Putting values of a,\(a_n , n\)

\(\Rightarrow \) 0 = –18 + (10 – 1) d

\(\Rightarrow \) 0 = -18 + 9d

\(\Rightarrow \) 18 = 9d

\(\Rightarrow \) d = 2

(iii) d = –3, n = 18, \(a_n = -5\)

We need to find a here.

Using formula \(a_n = a + (n - 1)d\)

Putting values of d, \(a_n , n\)

\(\Rightarrow \) –5 = a + (18 – 1) (–3)

\(\Rightarrow \) - 5 = a + (17) (-3)

\(\Rightarrow \) -5 = a – 51

\(\Rightarrow \) a = 46

(iv) a = –18.9, d = 2.5, \(a_n = 3.6\)

We need to find n here.

Using formula \(a_n = a + (n - 1)d\)

Putting values of \(d, a_n , n\)

\(\Rightarrow \) 3.6 = –18.9 + (n – 1) (2.5)

\(\Rightarrow \) 3.6 + 18.9 = (n - 1)(2.5)

\(\Rightarrow \) 22.5 = (n -1)(2.5)

\(\Rightarrow \) (n - 1) = \(\frac{22.5}{2.5}\)

\(\Rightarrow \) n - 1 = 9

\(\Rightarrow \) n = 10

(v) a = 3.5, d = 0, n = 105

We need to find \(a_n\) here.

Using formula \(a_n = a + (n - 1)d\)

Putting values of d, n and a,

\(\Rightarrow a_n\) = 3.5 + (105 - 1) (0)

\(\Rightarrow \)\(a_n = 3.5 + 104 × 0\)

\(\Rightarrow \) \(a_n = 3.5 + 0 \)

\(\Rightarrow a_n = 3.5\)

Q2 )
Choose the correct choice in the following and justify:

(i) 30th term of the AP : \(10, 7, 4…....\) is

(A) 97

(B) 77

(C) –77

(D) –87

(ii) 11th term of the AP : \(-3, \frac{1}{2}, 2…...\) is

(A) 28

(B) 22

(C) –38

(D) \({-48}{\dfrac{1}{2}}\)

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

(i) 10, 7, 4…

First term = a = 10,

Common difference = d

= 7 – 10 = 4 – 7 = –3

And n = 30{Because, we need to find 30th term}

\(a_n = a + (n - 1)d\)

= \(a_{30}\)

= 10 + (30 - 1) (-3)

= 10 – 87 = -77

Therefore, the answer is (C).

(ii) -3, \({{-1} \over {2}}\) , 2…

First term = a = –3,

Common difference = d

= \({{-1} \over {2}} - (-3) = 2 - ({{-1} \over {2}}) = {{5} \over {2}}\)

And n = 11 (Because, we need to find 11th term)

= -3 + (11 – 1) \({{5} \over {2}}\)

= -3 + 25 = 22

Therefore 11th term is 22 which means answer is (B).

Q3 )
In the following AP’s find the missing terms:

(i) 2, __ , 26

(ii) __, 13, __, 3

(iii) 5, __, __, \({9}{\dfrac{1}{2}}\)

(iv) –4. __, __, __, __, 6

(v) __, 38, __, __, __, –22

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

(i) 2, __ , 26

We know that difference between consecutive terms is equal in any A.P.

Let the missing term be x.

\(\Rightarrow \) x – 2 = 26 – x

\(\Rightarrow \) 2x = 28

\(\Rightarrow \) x = 14

Therefore, missing term is 14.

(ii)__, 13, __, 3

Let missing terms be x and y.

The sequence becomes x, 13, y, 3

We know that difference between consecutive terms is constant in any A.P.

\(\Rightarrow \) y – 13 = 3 – y

\(\Rightarrow \) 2y = 16

\(\Rightarrow \) y = 8

And 13 – x = y – 13

\(\Rightarrow \) x + y = 26

But, we have y = 8,

\(\Rightarrow \) x + 8 = 26

\(\Rightarrow \) x = 18

Therefore, missing terms are 18 and 8.

(iii) 5, __, __, \({9}{\dfrac{1}{2}}\)

Here, first term = a = 5 And,

4th term = \(a_4 = {9}{\dfrac{1}{2}}\)

Using formula \(a_n = a + (n - 1)d\) , to find nth term of arithmetic progression,

\(\Rightarrow a_4\) = 5 + (4 - 1) d

\(\Rightarrow {{19} \over {2}}\) = 5 + 3d

\(\Rightarrow \) 19 = 2 (5 + 3d)

\(\Rightarrow \) 19 = 10 + 6d

\(\Rightarrow \) 6d = 19 – 10

\(\Rightarrow \) 6d = 9

\(\Rightarrow \) d = \({{3} \over {2}}\)

Therefore, we get common difference

= d = \({{3} \over {2}}\)

Second term

= a + d = \(5 + {{3} \over {2}}

= {{13} \over {2}}\)

Third term = second term + d

= \({{13} \over {2}} + {{3} \over {2}} = {{16} \over {2}} = 8\)

Therefore, missing terms are \({{13} \over {2}}\) and 8

(iv)–4. __, __, __, __, 6

Here, First term = a = –4 and 6th term = \(a_6 = 6\)

Using formula \(a_n = a + (n - 1)d\) , to find nth term of arithmetic progression,

\(\Rightarrow a_6\) = -4 + (6 - 1) d

\(\Rightarrow \) 6 = -4 + 5d

\(\Rightarrow \) 5d = 10

\(\Rightarrow \) d = 2

Therefore, common difference = d = 2

Second term = first term + d

= a + d = –4 + 2 = –2

Third term = second term + d

= –2 + 2 = 0

Fourth term = third term + d

= 0 + 2 = 2

Fifth term = fourth term + d

= 2 + 2 = 4

Therefore, missing terms are –2, 0, 2 and 4.

(v) __, 38, __, __, __, –22

We are given 2nd and 6th term.

Using formula \(a_n = a + (n - 1)d\) , to find nth term of arithmetic progression,

\(a_2\) = a + (2 - 1) d And \(a_6 \) = a + (6 - 1) d

38 = a + d And -22 = a + 5d

These are equations in two variables, we can solve them using any method.

Using equation (38 = a + d), we can say that a = 38 - d.

Putting value of a in equation (-22 = a + 5d),

\(\Rightarrow\) -22 = 38 – d + 5d

\(\Rightarrow \) 4d = -60

\(\Rightarrow \) d = -15

Using this value of d and putting this in equation 38 = a + d,

38 = a – 15\(\Rightarrow \) a = 53

Therefore, we get a = 53 and d = -15

First term = a = 53

Third term = second term + d

= 38 – 15 = 23

Fourth term = third term + d

= 23 – 15 = 8

Fifth term = fourth term + d

= 8 – 15 = –7

Therefore, missing terms are 53, 23, 8 and –7.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

First term = a = 3,

Common difference = d = 8 – 3 = 13 – 8 = 5

\(a_n\) = 78

Using formula \(a_n = a + (n - 1)d\), to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow \) 78 = 3 + (n - 1) 5

\(\Rightarrow \) 75 = 5n - 5

\(\Rightarrow \) 80 = 5n

\(\Rightarrow \) n = 16

It means \(16^{th}\) term of the given AP is equal to 78.

Q5 )
Find the number of terms in each of the following APs:

(i) 7, 13, 19…., 205

(ii) 18,\({15}{\dfrac{1}{2}}\) , 13…, -47

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

(i) 7, 13, 19 …, 205

First term = a = 7,

Common difference = d

= 13 – 7 = 19 – 13 = 6

And \(a_n = 205\)

Using formula \(a_n = a + (n - 1)d\) , to find nth term of arithmetic progression,

\(\Rightarrow \) 205 = 7 + (n - 1) 6 = 7 + 6n – 6

\(\Rightarrow \) 205 = 6n + 1

\(\Rightarrow \) 204 = 6n

\(\Rightarrow \) n = 34

Therefore, there are 34 terms in the given arithmetic progression.

(ii) 18, \({15}{\dfrac{1}{2}}\) , 13 …, -47

First term = a =18,

Common difference = d

= \({15}{\dfrac{1}{2}} - 18 \)

\( = {{31} \over {2}} - 18 \)

\( = {{31 - 36} \over {2}} \)

\( = {{-5} \over {2}}\)

And \(a_n = -47\)

Using formula \(a_n = a + (n - 1)d\), to find nth term of arithmetic progression,

\(\Rightarrow \) 47 = 18 + (n - 1) \({{-5} \over {2}}\)

\(\Rightarrow \) 36 - \({{5} \over {2}}\) n + \({{5} \over {2}}\)

\(\Rightarrow\) -94 = 36 - 5n + 5

\(\Rightarrow \) 5n = 135

\(\Rightarrow \) n = 27

Therefore, there are 27 terms in the given arithmetic progression.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Let -150 is the \(n^{th}\) of AP 11, 8, 5, 2… which means that , \(a_n = -150\)

Here, First term = a = 11,

Common difference = d

= 8 – 11 = –3

Using formula \(a_n = a + (n - 1)d \), to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow\) -150 = 11 + (n - 1) (-3)

\(\Rightarrow \) -150 = 11 - 3n + 3

\(\Rightarrow \) 3n = 164

\(\Rightarrow \) n = \({{164} \over {3}}\)

But, n cannot be in fraction.

So, -150 is not a term of this A.P.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Given \(a_{11} = 38 \) and \(a_{16} = 73\)

Using formula \(a_n = a + ( n - 1)d \), to find nth term of arithmetic progression,

\(\Rightarrow \) 38 = a + (11 - 1)(d), and

73 = a + (16 - 1)(d)

\(\Rightarrow \) 38 = a + 10d, and

73 = a + 15d

These are equations consisting of two variables.

We have,

\(\Rightarrow \) 38 = a + 10d

\(\Rightarrow \) a = 38 - 10d

Let us put value of a in equation

\(\Rightarrow\) (73 = a + 15d),

\(\Rightarrow \) 73 = 38 - 10d + 15d

\(\Rightarrow \) 35 = 5d

Therefore, Common difference = d = 7

Putting value of d in equation

\(\Rightarrow \) 38 = a + 10d,

\(\Rightarrow \) 38 = a + 70

\(\Rightarrow \) a = -32

Therefore, common difference = d = 7 and

First term = a = –32

Using formula \(a_n = a + ( n - 1)d \), to find nth term of arithmetic progression,

\(\Rightarrow a_{31}\)= -32 + (31 - 1) (7)

\(\Rightarrow \) -32 + 210 = 178

Therefore, 31st term of AP is 178.

Q8 ) An AP consists of 50 terms of which \(3^{rd}\) term is 12 and the last term is 106. Find the \(29^{th}\) term.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

An AP consists of 50 terms and the \(50^{th}\) term is equal to 106 and \(a_3 = 12\)

Using formula \(a_n = a + (n - 1)d\) , to find nth term of arithmetic progression,

\(\Rightarrow a_{50}\)= a + (50 - 1)d, and

\(a_3\)= a + (3 - 1)d

\(\Rightarrow \) 106 = a + 49d and,

12 = a + 2d

These are equations consisting of two variables.

Using equation 106 = a + 49d,

we get a = 106 - 49d

Putting value of a in the equation 12 = a + 2d,

\(\Rightarrow \) 12 = 106 - 49d + 2d

\(\Rightarrow \) 47d = 94

\(\Rightarrow \) d = 2

Putting value of d in the equation,

a = 106 - 49d,

a = 106 – 49 (2)

= 106 – 98 = 8

Therefore, First term = a = 8 and

Common difference = d = 2

To find \(29^{th}\) term,

we use formula \(a_n = a + (n - 1)d\) which is used to find nth term of arithmetic progression,

\(a_{29}\)

= 8 + (29 - 1) 2

= 8 + 56 = 64

Therefore, 29th term of AP is equal to 64.

Q9 ) If the \(3^{rd}\) and the \(9^{th}\) terms of an AP are 4 and –8 respectively, which term of this AP is zero?

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

It is given that \(3^{rd}\) and \(9^{th}\) term of AP are 4 and –8 respectively.

It means \(a_3 = 4\) and \(a_9 = -8\)

Using formula \(a_n = a + (n - 1)d\), to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow \) 4 = a + (3 – 1)d and,

–8 = a + (9 – 1)d

\(\Rightarrow\) - 4 = a + 2d and,

-8 = a + 8d

These are equations in two variables.

Using equation 4 = a + 2d,

we can say that a = 4 - 2d

Putting value of a in other equation -8 = a + 8d,

\(\Rightarrow\) -8 = 4 - 2d + 8d

\(\Rightarrow\) -12 = 6d

\(\Rightarrow \) d = -2

Putting value of d in equation -8 = a + 8d,

\(\Rightarrow\) -8 = a + 8 (-2)

\(\Rightarrow \) -8 = a – 16

\(\Rightarrow \) a = 8

Therefore, first term = a = 8 and Common Difference = d = -2

We want to know which term is equal to zero.

Using formula \(a_n = a + (n - 1)d\) , to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow \) 0 = 8 + (n - 1) (-2)

\(\Rightarrow \) 0 = 8 - 2n + 2

\(\Rightarrow \) 0 = 10 - 2n

\(\Rightarrow \) 2n = 10

\(\Rightarrow \) n = 5

Therefore, \(5^{th}\) term is equal to 0.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

\( a_{17} = a_{10} + 7\)

Using formula \(a_n = a + (n - 1)d\), to find nth term of arithmetic progression,

\(a_{17}\) = a + 16d… (2)

\(a_{10}\) = a + 9d… (3)

Putting (2) and (3) in equation (1),

a + 16d = a + 9d + 7

\(\Rightarrow \) 7d = 7

\(\Rightarrow \) d = 1

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Lets first calculate \(54^{th}\) of the given AP.

First term = a = 3,

Common difference = d

= 15 – 3 = 12

Using formula \(a_n = a + (n - 1)d\), to find nth term of arithmetic progression,

\(a_{54}\) \)

\( = a + (54 - 1) d \)

\( = 3 + 53 (12) \)

= 3 + 636 = 639

We want to find which term is 132 more than its \(54^{th}\) term.

Let us suppose it is nth term which is 132 more than \(54^{th}\) term.

\(\Rightarrow a_n = a_{54} + 132\)

\(\Rightarrow \) 3 + (n - 1) 12 = 639 + 132

\(\Rightarrow \) 3 + 12n – 12 = 771

\(\Rightarrow \) 12n – 9 = 771

\(\Rightarrow \)12n = 780

\(\Rightarrow \) n = 65

Therefore, 65th term is 132 more than its \(54^{th}\) term.

Q12 ) Two AP’s have the same common difference. The difference between their \(100^{th}\) terms is 100, what is the difference between their \(1000^{th}\) terms.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Let first term of \(1^{st}\) AP = a

Let first term of \(2^{nd}\) AP = a'

It is given that their common difference is same.

Let their common difference be d.

It is given that difference between their \(100^{th}\) terms is 100.

Using formula \(a_n = a + (n - 1)d\), to find nth term of arithmetic progression,

\(\Rightarrow \) a + (100 - 1) d – [a' + (100 - 1) d]= 100

\(\Rightarrow \) a + 99d - a' - 99d = 100

\(\Rightarrow \) a - a' = 100… (1)

We want to find difference between their \(1000^{th}\) terms which means we want to calculate:

a + (1000 - 1) d – [a' + (1000 - 1) d]

= a + 999d - a' - 999d = a – a'

Putting equation (1) in the above equation,

a + (1000 - 1) d – [a' + (1000 - 1) d]

= a + 999d - a' + 999d = a - a' = 100

Therefore, difference between their \(1000^{th}\) terms would be equal to 100.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

We have AP starting from 105 because it is the first three digit number divisible by 7.

AP will end at 994 because it is the last three digit number divisible by 7.

Therefore, we have AP of the form 105, 112, 119…, 994

Let 994 is the \(n^{th}\) term of AP.

We need to find n here.

First term = a = 105,

Common difference = d

= 112 – 105 = 7

Using formula \(a_n = a + (n - 1)d\), to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow \) 994 = 105 + (n - 1) (7)

\(\Rightarrow \) 994 = 105 + 7n - 7

\(\Rightarrow \) 896 = 7n

\(\Rightarrow \) n = 128

It means 994 is the \(128^{th}\) term of AP.

Therefore, there are 128 terms in AP.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

First multiple of 4 which lie between 10 and 250 is 12.

The last multiple of 4 which lie between 10 and 250 is 248.

Therefore, AP is of the form 12, 16, 20… ,248

First term = a = 12,

Common difference = d = 4

Using formula \(a_n\) = a + (n – 1) d, to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow \) 248 = 12 + (n - 1) (4)

\(\Rightarrow \) 248 = 12 + 4n - 4

\(\Rightarrow \) 240 = 4n

\(\Rightarrow \) n = 60

It means that 248 is the \(60^{th}\) term of AP.

So, we can say that there are 60 multiples of 4 which lie between 10 and 250.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Lets first consider AP 63, 65, 67…

First term = a = 63,

Common difference = d

= 65 – 63 = 2

Using formula , to find \(n^{th}\) term of arithmetic progression,

\(a_n\) = 63 + (n - 1) (2)… (1)

Now, consider second AP 3, 10, 17…

First term = a = 3,

Common difference = d

= 10 – 3 = 7

Using formula , to find \(n^{th}\) term of arithmetic progression,

\(a_n\) = 3 + (n - 1) (7)… (2)

According to the given condition:

\(\Rightarrow\) (1) = (2)

\(\Rightarrow \) 63 + (n - 1) (2) = 3 + (n - 1) (7)

\(\Rightarrow \) 63 + 2n – 2 = 3 + 7n - 7

\(\Rightarrow \) 65 = 5n

\(\Rightarrow \) n = 13

Therefore, \(13^{th}\) terms of both the AP’s are equal.

Q16 ) Determine the AP whose third term is 16 and the \(7^{th}\) term exceeds the \(5^{th}\) term by 12.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Let first term of AP = a

Let common difference of AP = d

It is given that its 3rd term is equal to 16.

Using formula \(a_n = a + (n - 1)d\), to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow \) 16 = a + (3 - 1) (d)

\(\Rightarrow \) 16 = a + 2d… (1)

It is also given that \(7^{th}\) term exceeds \(5^{th}\) term by12.

According to the given condition:

\(\Rightarrow a_7 = a_5 + 12\)

\(\Rightarrow \) a + (7 – 1) d = a + (5 – 1) d + 12

\(\Rightarrow \) 2d = 12

\(\Rightarrow \) d = 6

Putting value of d in equation

\(\Rightarrow \) 16 = a + 2d,

\(\Rightarrow \) 16 = a + 2(6)

\(\Rightarrow \) a = 4

Therefore, first term = a = 4

And, common difference = d = 6

Therefore, AP is 4, 10, 16, 22…

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

We want to find \(20^{th}\) term from the last term of given AP.

So, let us write given AP in this way:253 … 13, 8, 3

Here First term = a = 253,

Common Difference = d

= 8 – 13 = –5

Using formula \(a_n = a + (n - 1)d\), to find \(n^{th}\) term of arithmetic progression,

= 253 + (20 - 1) (-5)

= 253 + 19 (-5)

= 253 – 95

= 158

Therefore, the \(20^{th}\) term from the last term of given AP is 158.

Q18 ) The sum of the \(4^{th}\) and \(8^{th}\) terms of an AP is 24 and the sum of \(6^{th}\) and \(10^{th}\) terms is 44. Find the three terms of the AP.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

The sum of\(4^{th}\) and \(8^{th}\) terms of an AP is 24 and sum of \(6^{th}\) and \(10^{th}\) terms is 44.

\(a_4 + a_8 = 24\) ,and \(a_6 + a_{10} = 44\)

Using formula \(a_n = a + (n - 1)d\), to find nth term of arithmetic progression,

\(\Rightarrow \) a + (4 - 1) d + [a + (8 - 1) d] = 24 and, a + (6 - 1) d + [a + (10 - 1) d] = 44

\(\Rightarrow \) a + 3d + a + 7d = 24, and a + 5d + a + 9d = 44

\(\Rightarrow \) 2a + 10d = 24, and 2a + 14d = 44

\(\Rightarrow \) a + 5d = 12, and a + 7d =22

These are equations in two variables.

Using equation, a + 5d = 12, we can say that a = 12 - 5d…...(1)

Putting (1) in equation a + 7d = 22,

\(\Rightarrow \) 12 - 5d + 7d = 22

\(\Rightarrow \) 12 + 2d = 22

\(\Rightarrow \) 2d = 10

\(\Rightarrow \) d = 5

Putting value of d in equation

a = 12 - 5d,

\(\Rightarrow \) a = 12 – 5 (5)

= 12 – 25

= -13

Therefore, first term = a = -13

and, Common difference = d = 5

Therefore, AP is –13, –8, –3, 2…

Its first three terms are –13, –8 and –3.

Q19 ) Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Subba Rao’s starting salary = Rs 5000

It means, first term = a = 5000

He gets an increment of Rs 200 after every year.

Therefore, common difference = d = 200

His salary after 1 year = 5000 + 200 = Rs 5200

His salary after two years = 5200 + 200 = Rs 5400

Therefore, it is an AP of the form 5000, 5200, 5400, 5600… , 7000

We want to know in which year his income reaches Rs 7000.

Using formula \(a_n = a + (n - 1)d\) , to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow \) 7000 = 5000 + (n - 1) (200)

\(\Rightarrow \) 7000 = 5000 + 200n - 200

\(\Rightarrow \) 7000 – 5000 + 200 = 200n

\(\Rightarrow \) 2200 = 200n

\(\Rightarrow \) n = 11

It means after 11 years, Subba Rao’s income would be Rs 7000.

Q20 ) Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Ramkali saved Rs. 5 in the first week of year. It means first term = a = 5

Ramkali increased her weekly savings by Rs 1.75.

Therefore, common difference = d = Rs 1.75

Money saved by Ramkali in the second week

= a + d = 5 + 1.75

= Rs 6.75

Money saved by Ramkali in the third week

= 6.75 + 1.75

= Rs 8.5

Therefore, it is an AP of the form:5, 6.75, 8.5 … , 20.75

We want to know in which year her weekly savings become 20.75.

Using formula \(a_n = a + (n - 1)d\), to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow \) 20.75 = 5 + (n - 1) (1.75)

\(\Rightarrow \) 20.75 = 5 + 1.75n - 1.75

\(\Rightarrow \)17.5 = 1.75n

\(\Rightarrow \) n = 10

It means in the \(10^{th}\) week her savings become Rs 20.75.

Q1 )
Find the sum of the following AP’s.

(i) 2, 7, 12… to 10 terms

(ii) –37, –33, –29… to 12 terms

(iii) 0.6, 1.7, 2.8… to 100 terms

(iv) \({{1} \over {15}},{{1} \over {12}},{{1} \over {10}}…\)to 11 terms

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

The formula for finding the sum of A.P. is \(S_n = {{n} \over {2}}[2a + (n - 1)d]\)

Where a= first term, d = common difference and n = number of terms

(i)2, 7, 12… to 10 terms

Here a = 2,

d = \(7 - 2 = 5\)

and n = 10

\(S_n = {{10} \over {2}}(4 + (10 - 1)5) \)

\(= 5(4 + 45) = 5(49) = 245\)

(ii) –37, –33, –29… to 12 terms

Here a = -37, d = \(-33 - (-37) = 4\) and n = 12

\(S_n = {{12} \over {2}}(-74 + (12 - 1)4) \)

\( = 6(-74 + 44) = 6(-30) = -180\)

(iii) 0.6, 1.7, 2.8… to 100 terms

Here a = 0.6, d = \(1.7 - 0.6 = 1.1\) and n = 100

\(S_n = {{100} \over {2}}(1.2 + (100 - 1)1.1) \)

\( = 50(1.2 + 108.9) \)

\( = 50(110.1) = 5505\)

(iv) \({{1} \over {15}},{{1} \over {12}},{{1} \over {10}}…\)to 11 terms

Here a = \({{1} \over {15}}\), d = \({{1} \over {12}} - {{1} \over {15}} = {{1} \over {60}}\) and n = 100

\(S_n = {{11} \over {2}}({{2} \over {15}} + (11 - 1){{1} \over {60}}) \)

\( = {{11} \over {2}}({{2} \over {15}} + {{1} \over {6}}) = {{11} \over {2}}({{9} \over {30}}) = {{33} \over {20}}\)

Q2 )
Find the sums given below:

(i) \(7 + {10}{\dfrac{1}{2}} + 14 + .......+ 84\)

(ii) 34 + 32 + 30 + .....… + 10

(iii) –5 + (–8) + (–11) + …..... + (–230)

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

(i) \(7 + {10}{\dfrac{1}{2}} + 14 +..........+ 84\)

Here First term = a = 7,

Common difference = d = \({{21} \over {2}} - 7 = {{21 - 15} \over {2}} = {{7} \over {2}} = 3.5\)

And Last term = l = 84

We do not know how many terms are there in the given AP.

So, we need to find n first.

Using formula \(a_n = a + (n - 1)d\) , to find nth term of arithmetic progression,

\(\Rightarrow\) [7 + (n - 1) (3.5)] = 84

\(\Rightarrow \) 7 + (3.5) n - 3.5 = 84

\(\Rightarrow \) 3.5n = 84 + 3.5 – 7

\(\Rightarrow \) 3.5n = 80.5

\(\Rightarrow \) n = 23

Therefore, there are 23 terms in the given AP.

It means n = 23.

Applying formula \(S_n = {{n} \over {2}} (a + l)\), to find sum of n terms of AP,

\(\Rightarrow S_n = {{23} \over {2}} (7 + 84)\)

\(\Rightarrow S_{23} = {{23} \over {2}} (91) = 1046.5\)

(ii) 34 + 32 + 30 + … + 10

Here First term = a = 34,

Common difference = d = 32 – 34 = –2

And Last term = l = 10

We do not know how many terms are there in the given AP.

So, we need to find n first.

Using formula \(a_n = a + (n - 1)d\), to find nth term of arithmetic progression,

\(\Rightarrow\) [34 + (n - 1) (-2)] = 10

\(\Rightarrow \) 34 – 2n + 2 = 10

\(\Rightarrow \) 2n = 26

\(\Rightarrow \) n = 13

Therefore, there are 13 terms in the given AP.

It means n = 13.

Applying formula \(S_n = {{n} \over {2}} (a + l)\), to find sum of n terms of AP,

\( S_{13} = {{13} \over {2}} (34 + 10) \)

\(= {{13} \over {2}} (44) = 286\)

(iii) -5 + (-8) + (-11) + … + (-230)

Here First term = a = –5,

Common difference = d

= –8 – (–5) = –8 + 5 = –3

And Last term = l = -230

We do not know how many terms are there in the given AP.

So, we need to find n first.

Using formula \(a_n = a + (n - 1)d\), to find nth term of arithmetic progression,

\(\Rightarrow\) [-5 + (n - 1) (-3)] = -230

\(\Rightarrow \) -5 - 3n + 3 = -230

\(\Rightarrow\) -3n = -228

\(\Rightarrow \) n = 76

Therefore, there are 76 terms in the given AP.

It means n = 76.

Applying formula \(S_n = {{n} \over {2}} (a + l)\), to find sum of n terms of AP,

\(S_{76} = {{76} \over {2}} (-5 - 230) \)

\( = 38 (-235) = -8930\)

Q3 )
In an AP

(i)Given a = 5, d = 3, an = 50, find n and \(S_n\).

(ii) Given a = 7, \(a_{13}\) = 35, find d and \(S_{13}\).

(iii) Given \(a_{12}\) = 37, d = 3, find a and \(S_{12}\).

(iv) Given \(a_3\) = 15, \(S_{10}\) = 125, find d and \(a_{10}\).

(v) Given d = 5, \(S_9\)= 75, find a and \(a_9\).

(vi) Given a = 2, d = 8, \(S_n\) = 90, find n and an.

(vii) Given a = 8, an = 62, \(S_n\) = 210, find n and d.

(viii) Given \(a_n\) = 4, d = 2, \(S_n\) = - 14, find n and a.

(ix) Given a = 3, n = 8, S = 192, find d.

(x)Given l = 28, S = 144 and there are total 9 terms. Find a.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

(i) Given that, a = 5, d = 3, \(a_n\) = 50

As we know, from the formula of the nth term in an AP,

\(a_n = a + (n – 1)d\)

Therefore, putting the given values, we get,

\(\Rightarrow \) 50 = 5+(n -1)×3

\(\Rightarrow \) 3(n -1) = 45

\(\Rightarrow \) n -1 = 15

\(\Rightarrow \) n = 16

Now, sum of n terms,

\(\Rightarrow S_n = (a + a_n)\)

\(\Rightarrow S_n = {{16} \over{2}} (5 + 50) = 440\)

(ii) Given that, a = 7, \(a_{13}\) = 35

As we know, from the formula of the nth term in an AP

\(a_n = a + (n – 1)d\)

Therefore, putting the given values, we get,

\(\Rightarrow \) 35 = 7+(13-1)d

\(\Rightarrow \) 12d = 28

\(\Rightarrow \) d = 28/12 = 2.33

Now, sum of n terms,

\(\Rightarrow S_n = (a + a_n)\)

\(\Rightarrow S_n = {{13} \over {2}} (7 + 35) = 273\)

(iii)Given that, \(a_{12}\) = 37, d = 3

As we know, from the formula of the nth term in an AP,

\(a_n = a + (n – 1)d\)

Therefore, putting the given values, we get,

\(\Rightarrow a_{12}\) = a + (12 - 1)3

\(\Rightarrow \) 37 = a + 33

\(\Rightarrow \) a = 4

Now, sum of nth term,

\(\Rightarrow S_n = (a + a_n)\)

\(\Rightarrow S_n = {{12} \over {2}} (4 + 37)\)

= 246

(iv) Given that, \(a_3\) = 15, \(S_{10}\) = 12

As we know, from the formula of the nth term in an AP,

\(a_n = a + (n – 1)d\)

Therefore, putting the given values, we get,

\(\Rightarrow a_3\) = a+(3 - 1)d

\(\Rightarrow\) 15 = a + 2d ………………………….. (i)

Sum of the nth term,

\(\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]\)

\(\Rightarrow S_{10} = {{10} \over {2}} [2a+(10-1)d] \)

\(\Rightarrow \) 125 = 5(2a+9d)

\(\Rightarrow \) 25 = 2a+9d ……………………….. (ii)

On multiplying equation (i) by 2, we will get;

30 = 2a+4d ………………………………. (iii)

By subtracting equation (iii) from (ii), we get,

-5 = 5d

d = -1

From equation (i)

15 = a + 2(-1)

15 = a - 2

a = 17 = First term

\(\Rightarrow a_{10}\) = a + (10-1)d

\(\Rightarrow a_{10}\) = 17 + (9)(-1)

\(\Rightarrow a_{10}\) = 17 -9 = 8

(v) Given that, d = 5, \(S_9\) = 75

As, sum of n terms in AP is,

\( S_n = {{n} \over {2}} [2a+(n-1)d]\)

Therefore, the sum of first nine terms are;

\(\Rightarrow S_9 = {{9} \over {2}} [2a+(9 -1)5]\)

\(\Rightarrow \) =25 = 3(a+20)

\(\Rightarrow \) 25 = 3a+60

\(\Rightarrow \) 3a = 25-60

\(\Rightarrow \) a = \({{-35} \over {3}}\)

As we know, the nth term can be written as;

\(a_n = a + (n – 1)d\)

=\(a_9 = a + (9 - 1)(5)\)

= \({{-35} \over {3}}\) +8(5)

= \({{-35} \over {3}}\)+40

= (\({{-35 + 120} \over {3}}\))

= \({{85} \over {3}}\)

(vi) Given that, a = 2, d = 8, \(S_n\) = 90

As, sum of n terms in an AP is,

\(\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]\)

\(\Rightarrow 90 = {{n} \over {2}} [2a +(n -1)d]\)

\(\Rightarrow 180 = n(4 +8n - 8)\)

\(\Rightarrow 180 = n(8n-4) = 8n^2-4n\)

\(\Rightarrow 8n^2 - 4n –180 = 0\)

\(\Rightarrow 2n^2 - n - 45 = 0\)

\(\Rightarrow 2n^2 -10n + 9n - 45 = 0\)

\(\Rightarrow 2n(n - 5) + 9(n - 5) = 0\)

\(\Rightarrow(2n - 9)(2n + 9) = 0\)

So, n = 5 (as it is positive integer)

\(a_5\) = 8 + 5×4 = 34

(vii) Given that, a = 8, \(a_n\) = 62, \(S_n\) = 210

As, sum of n terms in an AP is,

\(\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]\)

\(\Rightarrow 210 = {{n} \over {2}} (8 +62)\)

\(\Rightarrow \) 35n = 210

\(\Rightarrow \) n = \(\frac{210}{35}\) = 6

Now, 62 = 8 + 5d

\(\Rightarrow \) 5d = 62 - 8 = 54

\(\Rightarrow \) d = \({{54} \over {5}}\) = 10.8

(viii) Given that, nth term, an = 4,

common difference, d = 2, sum of n terms, \(S_n\) = -14.

As we know, from the formula of the nth term in an AP,

\( a_n = a + (n – 1)d\)

Therefore, putting the given values, we get,

\(\Rightarrow \) 4 = a+(n -1)2

\(\Rightarrow \) 4 = a+2n-2

\(\Rightarrow \) a+2n = 6

\(\Rightarrow \) a = 6 - 2n ……………………. (i)

As we know, the sum of n terms is;

\(\Rightarrow S_n = \frac{n}{2} (a + a_n) \)

\(\Rightarrow -14 = \frac{n}{2}(a+4)\)

\(\Rightarrow-28 = n(a + 4)\)

\(\Rightarrow-28 = n(6 -2n + 4) \){From equation (i)}

\(\Rightarrow-28 = n(-2n +10)\)

\(\Rightarrow-28 = - 2n^2 + 10n\)

\(\Rightarrow 2n^2 -10n - 28 = 0\)

\(\Rightarrow n^2 -5n -14 = 0\)

\(\Rightarrow n^2 -7n + 2n -14 = 0\)

\(\Rightarrow n(n - 7) + 2(n - 7) = 0\)

\(\Rightarrow (n -7)(n +2) = 0\)

Either n - 7 = 0 or n + 2 = 0

n = 7 or n = -2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we get

\(\Rightarrow \) a = 6 - 2n

\(\Rightarrow \) a = 6 - 2(7)

\(\Rightarrow\) = 6 - 14

Thus, a = -8

(ix) Given that, first term, a = 3,

Number of terms, n = 8

And sum of n terms, S = 192

As we know,

\(\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]\)

\( \Rightarrow 192 = {{8} \over {2}} [2×3+(8 -1)d]\)

\(\Rightarrow \) 192 = 4[6 + 7d]

\(\Rightarrow \) 48 = 6 + 7d

\(\Rightarrow \) 42 = 7d

\(\Rightarrow \) d = 6

(x) Given that, l = 28,S = 144 and there are total of 9 terms.

Sum of n terms formula,

\(\Rightarrow S_n = {{n} \over {2}} (a + l)\)

\(\Rightarrow 144 = {{9} \over {2}}(a+28)\)

\(\Rightarrow \) (16)×(2) = a + 28

\(\Rightarrow \) 32 = a + 28

\(\Rightarrow \) a = 4

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Let there be n terms of the AP. 9, 17, 25 …

For this A.P.,

First term, a = 9

Common difference, d

= \(a_2 – a_1\) = 17-9 = 8

As, the sum of n terms, is;

\(\Rightarrow S_n = {{n} \over {2}} (2a + (n – 1)d\)

\(\Rightarrow \) 636 = \({{n} \over {2}}\) [2×a+(8-1)×8]

\(\Rightarrow \) 636 = \({{n} \over {2}}\) [18+(n-1)×8]

\(\Rightarrow \) 636 = n [9 +4n -4]

\(\Rightarrow \) 636 = n (4n +5)

\(\Rightarrow \) 4\(n^2\) +5n -636 = 0

\(\Rightarrow \) 4\(n^2\) +53n -48n -636 = 0

\(\Rightarrow \) n (4n + 53)-12 (4n + 53) = 0

\(\Rightarrow\) (4n +53)(n -12) = 0

Either 4n+53 = 0 or n-12 = 0

n = \({{-53} \over {4}}\) or n = 12

n cannot be negative or fraction, therefore, n = 12 only.

Q5 ) The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Given that,

first term, a = 5

last term, l = 45

Sum of the AP, \(S_n\) = 400

As we know, the sum of AP formula is;

\(\Rightarrow S_n = {{n} \over {2}} (a+l)\)

\(\Rightarrow \) 400 = \({{n} \over {2}}\)(5+45)

\(\Rightarrow \) 400 = \({{n} \over {2}}\)(50)

Number of terms, n =16

As we know, the last term of AP series can be written as;

l = a+(n -1)d

\(\Rightarrow \) 45 = 5 +(16 -1)d

\(\Rightarrow \) 40 = 15d

\(\Rightarrow \) Common difference, d = \({{40} \over {15}} = {{8} \over {3}}\)

Q6 ) The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Given that,

First term, a = 17

Last term, l = 350

Common difference, d = 9

Let there be n terms in the A.P., thus the formula for last term can be written as;

\(\Rightarrow \) l = a+(n - 1)d

\(\Rightarrow \) 350 = 17+(n - 1)9

\(\Rightarrow \) 333 = (n -1)9

\(\Rightarrow\) (n- 1) = 37

\(\Rightarrow \) n = 38

\(\Rightarrow S_n = {{n} \over {2}} (a+l)\)

\(\Rightarrow S_{38} = {{13} \over {2}} (17+350)\)

= 19×367

= 6973

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Given,

Common difference, d = 7

\(22^{nd}\) term, \(a_{22}\) = 149

We have to find the sum of first 22 term, \(S_{22}\)

By the formula of nth term,

\(\Rightarrow a_n = a + (n - 1)d\)

\(\Rightarrow a_{22} = a + (22 - 1)d\)

\(\Rightarrow \) 149 = a + (21)(7)

\(\Rightarrow \) 149 = a + 147

\(\Rightarrow \) a = 2 = First term

Sum of n terms,

\(\Rightarrow S_n = {{n} \over {2}}(a+a_n)\)
= \({{22} \over {2}}\) (2+149)
= 11×151
= 1661

Q8 ) Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Given that,

Second term, \(a_2\) = 14

Third term, \(a_3\) = 18

Common difference, d

= \(a_3 - a_2\) = 18-14 = 4

\(a_2\) = a + d

14 = a + 4

a = 10 = First term

Sum of n terms;

\(S_n = {{n} \over {2}} [2a + (n – 1)d]\)

\(S_{51} = {{51} \over {2}} [2×10 (51-1) 4]\)

= \({{51} \over {2}}[2 + (20)4\)]

= 51×\({{220} \over {2}}\)

= 51×110

= 5610

Q9 ) If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Given that,

\(S_7\) = 49

\(S_{17}\) = 289

We know, Sum of nth term;

\(\Rightarrow S_n = {{n} \over {2}} [2a + (n – 1)d]\)

\(\Rightarrow S_7= {{7} \over {2}} [2a + (n -1)d]\)

\(\Rightarrow S_7 = {{7} \over {2}} [2a + (7 -1)d]\)

\(\Rightarrow 49 ={{7} \over {2}} [2a + 6d]\)

\(\Rightarrow \) 7 = (a+3d)

\(\Rightarrow \) a + 3d = 7 …........ (i)

In the same way,

\(\Rightarrow S_{17} = {{17} \over {2}} [2a + (17-1)d]\)

\(\Rightarrow 289 = {{17} \over {2}} (2a + 16d)\)

\(\Rightarrow \) 17 = (a + 8d)

\(\Rightarrow \) a + 8d = 17 ……… (ii)

Subtracting equation (i) from equation (ii),

5d = 10

d = 2

From equation (i), we can write it as;

\(\Rightarrow \) a + 3(2) = 7

\(\Rightarrow \) a + 6 = 7

\(\Rightarrow \) a = 1

Hence,

\(\Rightarrow S_n = {{n} \over {2}} [2a + (n – 1)d]\)

= \({{n} \over {2}}[2(1)+(n – 1)×2]\)
= \({{n} \over {2}}(2+2n-2)\)
= \({{n} \over {2}}(2n)\)
= \(n^2\)

Q10 )
Show that \(a_1, a_2 … , a_n , …\) form an AP where an is defined as below

(i) \(a_n = 3+4_n\)

(ii) \(a_n = 9 - 5_n\)

Also find the sum of the first 15 terms in each case.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

(i) \(\because a_n = 3+4_n\)

\( a_1 = 3+4(1) = 7\)

\( a_2 = 3+4(2) = 3+8 = 11\)

\( a_3 = 3+4(3) = 3+12 = 15\)

\( a_4 = 3+4(4) = 3+16 = 19\)

We can see here, the common difference between the terms are;

\( a_2 - a_1 = 11-7 = 4\)

\( a_3 - a_2 = 15-11 = 4\)

\( a_4 - a_3 = 19-15 = 4\)

Hence, \(a_k + 1 - a_k\) is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.

Now, we know, the sum of \(n^{th}\) term is;

\(S_n = {{n} \over {2}}[2a+(n -1)d]\)

\(\therefore S_{15} = {{15} \over {2}}[2(7)+(15-1)×4]\)

= \({{15} \over {2}}[(14)+56]\)

= \({{15} \over {2}}(70)\)

= 15×35

= 525

(ii) \(a_n = 9 - 5n\)

\(a_1 = 9-5×1 = 9-5 = 4\)

\(a_2 = 9-5×2 = 9-10 = -1\)

\(a_3 = 9-5×3 = 9-15 = -6\)

\(a_4 = 9-5×4 = 9-20 = -11\)

We can see here, the common difference between the terms are;

\(a_2 - a_1 = -1-4 = -5\)

\(a_3 - a_2 = -6-(-1) = -5\)

\(a_4 - a_3 = -11-(-6) = -5\)

Hence, \(a_k + 1 - a_k\) is same every time. Therefore, this is an A.P. with common difference as -5 and first term as 4.

Now, we know, the sum of nth term is;

\(S_n = {{n} \over {2}} [2a +(n-1)d]\)

\(S_{15} = {{15} \over {2}}[2(4) +(15 -1)(-5)]\)

= \({{15} \over {2}}\)[8 + 14(-5)]

= \({{15} \over {2}}\)(8 - 70)

= \({{15} \over {2}}\)(-62)

= 15(-31)

= -465

Q11 ) If the sum of the first n terms of an AP is \(4n - n^2\), what is the first term (that is \(S_1\))? What is the sum of first two terms? What is the second term? Similarly find the \(3^{rd}\), the \(10^{th}\) and the nth terms.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Given that,

\(S_n = 4n-n^2\)

First term, a = \(S_1 = 4(1) - (1)2 = 4-1 = 3\)

Sum of first two terms = \(S_2= 4(2)-(2)2 = 8-4 = 4\)

Second term, \(a_2 = S_2 - S_1 = 4-3 = 1\)

Common difference, d = \(a_2-a = 1-3 = -2\)

Nth term, \(a_n = a+(n-1)d \)

= 3 + (n -1)(-2)

= 3 - 2n + 2

= 5 - 2n

Therefore,\( a_3 = 5 - 2(3) = 5 - 6 = -1\)

\(a_{10} = 5-2(10) = 5 - 20 = -15\)

Hence, the sum of first two terms is 4. The second term is 1.

The 3rd, the 10th, and the nth terms are -1, -15, and 5 - 2n respectively.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

The positive integers that are divisible by 6 are 6, 12, 18, 24 ….

We can see here, that this series forms an A.P. whose first term is 6 and common difference is 6.

a = 6

d = 6

We have to find \(S_{40}\)

By the formula of sum of n terms, we know,

\(S_n = {{n} \over {2}} [2a +(n – 1)d]\)

Therefore, putting n = 40, we get,

\(S_{40} = {{40} \over {2}} [2(6)+(40-1)6]\)

= \(20[12 + (39)(6)]\)

= \(20(12 + 234)\)

= \(20 × 246\)

= \(4920\)

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

The multiples of 8 are 8, 16, 24, 32…

The series is in the form of AP, having first term as 8 and common difference as 8.

Therefore, a = 8

d = 8

We have to find \(S_{15}\)

By the formula of sum of \(n^{th}\) term, we know,

\(S_n = {{n} \over {2}} [2a+(n-1)d]\)

\(S_{15} = {{15} \over {2}} [2(8) + (15-1)8] \)

\(= {{15} \over {2}}[6 +(14)(8)] \)

\( = {{15} \over {2}}[16 +112]\)

\(= {{15(128)} \over {2}}\)

\(= 15 × 64\)

\(= 960\)

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.

Therefore, we can see that these odd numbers are in the form of A.P.

Hence,

First term, a = 1

Common difference, d = 2

Last term, l = 49

By the formula of last term, we know,

\(l = a+(n-1) d \)

\(\Rightarrow 49 = 1+(n-1)2 \)

\(\Rightarrow 48 = 2(n -1)\)

\(\Rightarrow n - 1 = 24 \)

\(\Rightarrow \) n = 25 = Number of terms

By the formula of sum of \(n^{th}\) term, we know,

\(S_n = {{n} \over {2}}(a +l)\)

\(S_{25} = {{25} \over {2}} (1+49) \)

\(= {{25(50)} \over {2}} \)

\(=(25)(25) \)

\(= 625\)

Q15 ) A contract on construction job specifies a penalty for delay of completion beyond a certain dates follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

We can see, that the given penalties are in the form of A.P. having first term as 200 and common difference as 50.

Therefore, a = 200 and d = 50

Penalty that has to be paid if contractor has delayed the work by 30 days = \(S_{30}\)

By the formula of sum of nth term, we know,

\(S_n = {{n} \over {2}}[2a+(n -1)d]\)

\(S_{30}= {{30} \over {2}}[2(200)+(30 – 1)50] \)

= 15[400+1450]

= 15(1850)

= 27750

Therefore, the contractor has to pay Rs 27750 as penalty.

Q16 ) A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Let the cost of 1st prize be P Rs.

Cost of 2nd prize = P - 20 Rs

And cost of 3rd prize = P - 40 Rs

We can see that the cost of these prizes are in the form of A.P., having common difference as -20 and first term as P.

Thus, a = P and d = -20

Given that, \(S_7 = 700\)

By the formula of sum of nth term, we know,

\(\Rightarrow S_n = {{n} \over {2}} [2a + (n – 1)d]\)

\(\Rightarrow{{7} \over {2}} [2a + (7 – 1)d] = 700\)

\(\Rightarrow{{2a + (6)(-20)} \over {2}} = 100\)

\(\Rightarrow \) a + 3(-20) = 100

\(\Rightarrow \) a - 60 = 100

\(\Rightarrow \) a = 160

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

Q17 ) In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

t can be observed that the number of trees planted by the students is in an AP.

1, 2, 3, 4, 5………………..12

First term, a = 1

Common difference, d = 2 - 1 = 1

\(S_n = {{n} \over {2}} [2a +(n-1)d]\)

\(S_{12} = {{12} \over {2}} [2(1)+(12-1)(1)] \)

= 6(2+11)

= 6(13)

= 78

Therefore, number of trees planted by 1 section of the classes = 78

Number of trees planted by 3 sections of the classes = 3×78 = 234

Therefore, 234 trees will be planted by the students.

Q18 ) A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take \(\pi = \frac{22}{7} \) )

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

We know,

Perimeter of a semi-circle = \(\pi\)r

Therefore,

\(P_1 = \pi(0.5) = {{\pi} \over {2}}\) cm

\(P_2 = \pi(1) = \pi\) cm

\(P_3 = \pi(1.5) = {{3pi} \over { 2}}\) cm

Where, \(P_1, P_2, P_3\) are the lengths of the semi-circles.

Hence we got a series here, as,

\({{\pi} \over {2}} , \pi, {{3\pi} \over { 2}}, 2\pi,\) ….

\(P_1 = {{\pi} \over {2}}\) cm

\(P_2 = \pi\) cm

Common difference, \(d = P_2- P_1 = \pi – {{\pi} \over {2}} = {{\pi} \over {2}}\)

First term = \(P_1= a = {{\pi} \over {2}}\) cm

By the sum of n term formula, we know,

\(S_n = {{n} \over {2}} [2a + (n – 1)d]\)

Therefor, Sum of the length of 13 consecutive circles is;

\(S_{13} = {{13} \over {2}} [2({{\pi} \over {2}}) + (13 – 1){{\pi} \over {2}}]\)

= \( \frac{13}{2} [\pi + 6\pi]\)

=\(\frac{13}{2} (7\pi)\)

= \(\frac{13}{2} × 7 × \frac{22}{7}\)

= \(143\) cm

Q19 ) 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

We can see that the numbers of logs in rows are in the form of an A.P.20, 19, 18…

For the given A.P.,

First term, a = 20

and common difference,d

\(= a_2 - a_1 = 19-20 = -1\)

Let a total of 200 logs be placed in n rows.

Thus, \(S_n = 200\)

By the sum of \(n^{th}\) term formula,

\(\Rightarrow S_n = {{n} \over {2}} [2a +(n -1)d]\)

\(\Rightarrow S_{12} = {{12} \over {2}} [2(20)+(n -1)(-1)]\)

\(\Rightarrow \) 400 = n (40- n+1)

\(\Rightarrow \) 400 = n (41-n)

\(\Rightarrow 400 = 41n-n^2\)

\(\Rightarrow n^2-41n + 400 = 0\)

\(\Rightarrow n^2-16n-25n+400 = 0\)

\(\Rightarrow \) n(n -16)-25(n -16) = 0

\(\Rightarrow\) (n -16)(n -25) = 0

Either (n -16) = 0 or n-25 = 0

\(\Rightarrow \) n = 16 or n = 25

By the nth term formula,

\(\Rightarrow a_n = a+(n-1)d\)

\(\Rightarrow a_{16} = 20+(16-1)(-1)\)

\(\Rightarrow a_{16} = 20-15\)

\(\Rightarrow a_{16} = 5\)

Similarly, the 25th term could be written as;

\(\Rightarrow a_{25} = 20+(25-1)(-1)\)

\(\Rightarrow a_{25} = 20-24\)

\(\Rightarrow a_{25}= -4 \)

It can be seen, the number of logs in 16th row is 5 as the numbers cannot be negative.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.

Q20 )
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2×5+2×(5+3)]

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

The distances of potatoes from the bucket are 5, 8, 11, 14…, which is in the form of AP.

Given, the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.

Therefore, distances to be run w.r.t distances of potatoes, could be written as;

10, 16, 22, 28, 34,……….

Hence, the first term, a = 10 and d = 16 - 10 = 6

S10 =?

By the formula of sum of n terms, we know,

\(S_{10} = {{12} \over {2}} [2(20)+(n -1)(-1)]\)

= 5[20+54]

= 5(74)

= 370

Therefore, the competitor will run a total distance of 370 m.

Q1 ) Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find n for an < 0]

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Given the AP series is 121, 117, 113, . . .,

Thus, first term, a = 121

Common difference, d = 117-121= -4

By the \(n^{th}\) term formula,

\(a_n = a+(n -1)d\)

Therefore,

\(a_n = 121+(n-1)(-4)\)

= 121-4n+4

=125-4n

To find the first negative term of the series, \(a_n < 0\)

Therefore,

\(\Rightarrow \) 125-4n < 0

\(\Rightarrow \) 125 < 4n

\(\Rightarrow \) n>125/4

\(\Rightarrow \) n>31.25

Therefore, the first negative term of the series is \(32^{nd}\) term.

Q2 ) The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

From the given statements, we can write,

\(a_3 + a_7 = 6\) …………………………….(i)

And

\(a_3 ×a_7 = 8\) ……………………………..(ii)

By the \(n^{th}\) term formula,

\(a_n = a+(n - 1)d\)

Third term, \(a_3 = a+(3 -1)d\)

\(a_3 = a + 2d\)………………………………(iii)

And Seventh term, \(a_7= a+(7-1)d\)

\(a_7 = a + 6d \)………………………………..(iv)

From equation (iii) and (iv), putting in equation(i), we get,

\(\Rightarrow \) a+2d +a+6d = 6

\(\Rightarrow \) 2a+8d = 6

\(\Rightarrow \) a+4d=3

\(\Rightarrow \) a = 3–4d …………………………………(v)

Again putting the eq.(iii) and (iv), in eq. (ii), we get,

(a+2d)×(a+6d) = 8

Putting the value of a from equation (v), we get,

\(\Rightarrow (3–4d +2d)×(3–4d+6d) = 8\)

\(\Rightarrow (3 –2d)×(3+2d) = 8\)

\(\Rightarrow 3^2 – 2d^2\) = 8

\(\Rightarrow 9 – 4d^2\) = 8

\(\Rightarrow 4d^2 = 1\)

\(\Rightarrow d = {{1} \over {2}} or -{{1} \over {2}}\)

Now, by putting both the values of d, we get,

\(\Rightarrow a = 3 – 4d = 3 – 4({{1} \over {2}}) = 3 – 2 = 1\), when \(d = {{1} \over {2}}\)

\(a = 3 – 4d = 3 – 4(-{{1} \over {2}}) = 3+2 = 5\), when \(d = -{{1} \over {2}}\)

We know, the sum of nth term of AP is;

\(S_n = {{n} \over {2}} [2a +(n – 1)d]\)

So, when a = 1 and d=\({{1} \over {2}}\)

Then, the sum of first 16 terms are;

\(S_{16} = {{16} \over {2}} [2 +(16-1){{1} \over {2}}] \)

\(= 8(2+{{15} \over {2}}) = 76\)

And when a = 5 and d= -\({{1} \over {2}}\)

Then, the sum of first 16 terms are;

\(S_{16} = {{16} \over {2}} [2.5+(16-1)(-{{1} \over {2}})] \)

\( = 8({{5} \over {2}})=20\)

Q3 ) A ladder has rungs 25 cm apart.. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \({2} {\dfrac{1}{2}}\) apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = -250/25 ].

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Given,

Distance between the rungs of the ladder is 25 cm.

Distance between the top rung and bottom rung of the ladder is =

\({2} {\dfrac{1}{2}}m \)

\( = {2} {\dfrac{1}{2}} (100)cm \)

\( = {{5} \over {2}} ×100\) cm

= 250 cm

Therefore, total number of rungs = \({{250} \over {25}} + 1 = 11\)

As we can see from the figure, the ladder has rungs in decreasing order from top to bottom. Thus, we can conclude now, that the rungs are decreasing in an order of AP.

And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.

So,

First term, a = 45

Last term, l = 25

Number of terms, n = 11

Now, as we know, sum of nth terms is equal to,

\(S_n= {{n} \over {2}}(a+ l)\)

\(S_n= {{11} \over {2}}(45+25) \)

\( = {{11} \over {2}}(70) = 385\) cm

Hence, the length of the wood required for the rungs is 385 cm.

Q4 )
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

[Hint :Sx – 1 = S49 – Sx]

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

Given,

Row houses are numbers from 1,2,3,4,5…….49.

Thus we can see the houses numbered in a row are in the form of AP.

So,

First term, a = 1

Common difference, d=1

Let us say the number of \(x^{th}\) houses can be represented as;

Sum of nth term of AP = \({{n} \over {2}}[2a + (n - 1)d]\)

Sum of number of houses beyond x house = Sx-1

= \({{x - 1} \over {2}}[2(1) + (x -1-1)1]\)

= \({{x - 1} \over {2}} [2+x-2]\)

= \({{x(x - 1)} \over {2}}\)………………………………(i)

By the given condition, we can write,

\(S_{49} – Sx = ({{49} \over {2}}[2.1+(49-1)1])–({{x} \over {2}}[2.1+(x-1)1])\)

= \(25(49) – {{x( x + 1)} \over {2}} \)………………………………….(ii)

As per the given condition, eq.(i) and eq(ii) are equal to each other;

Therefore,

\({{x(x - 1)} \over {2}} = 25(49) – {{x ( x - 1)} \over {2}}\)

\(x = ±35\)

As we know, the number of house cannot be an a negative number. Hence, the value of x is 35

Q5 ) A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \({{1} \over {4}}\) m and a tread of \({{1} \over {2}}\) m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = \({{1} \over {4}} × {{1} \over {2}} × 50 m^3\).]

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Answer :

As we can see from the given figure, the first step is \({{1} \over {2}}\) m wide, 2nd step is 1 m wide and 3rd step is \({{3} \over {2}}\) m wide. Thus we can understand that the width of step by \({{1} \over {2}}\) m each time when height is \({{1} \over {4}}\) m. And also, given length of the steps is 50 m all the time. So, the width of steps forms a series AP in such a way that;

\({{1} \over {2}}\) , 1, \({{3} \over {2}}\), 2, ……..

Volume of steps = Volume of Cuboid

= Length × Breadth Height

Now,

Volume of concrete required to build the first step = \({{1} \over {4}} × {{1} \over {2}} ×50 = {{25} \over {4}}\)

Volume of concrete required to build the second step = \({{1} \over {5}} × {{2} \over {2}} × 50 = {{25} \over {2}}\)

Volume of concrete required to build the second step = \({{1} \over {4}} × {{3} \over {2}}×50 = {{75} \over {2}}\)

Now, we can see the volumes of concrete required to build the steps, are in AP series;

\({{25} \over {4}} , {{25} \over {2}} , {{75} \over {2}} \)…..

Thus, applying the AP series concept,

First term, a = \({{25} \over {4}}\)

Common difference, d = \({{25} \over {2}} – {{25} \over {4}} = {{25} \over {4}}\)

As we know, the sum of n terms is;

\(S_n = {{n} \over {2}}[2a+(n-1)d] \)

\(= {{15} \over {2}}(2×({{25} \over {4}} )+({{15} \over {2}} -1){{25} \over {4}})\)

Upon solving, we get,

\(S_n = {{15} \over {2}} (100)\)

\(S_n= 750\)

Hence, the total volume of concrete required to build the terrace is \(750 m^3\).

There are total 49 questions present in ncert solutions for class 10 maths chapter 5 arithmetic progressions

There are total 19 long question/answers in ncert solutions for class 10 maths chapter 5 arithmetic progressions

There are total 4 exercise present in ncert solutions for class 10 maths chapter 5 arithmetic progressions