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(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) \(1^2, 5^2, 7^2, 73\)…

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

- 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.
- 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
- 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 …

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