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(i) 2, __ , 26

(ii) __, 13, __, 3

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

(iv) –4. __, __, __, __, 6

(v) __, 38, __, __, __, –22

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.

- 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) \(a = 3.5, d = 0, n = 105\)
- 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) \({-48}{\dfrac{1}{2}}\)
- Which term of the AP: 3, 8, 13, 18 … is 78?
- Find the number of terms in each of the following APs: (i) 7, 13, 19…., 205 (ii) 18,\({15}{\dfrac{1}{2}}\) , 13…, -47
- Check whether –150 is a term of the AP: 11, 8, 5, 2…
- Find the 31st term of an AP whose \(11^{th}\) term is 38 and \(16^{th}\) term is 73.
- An AP consists of 50 terms of which \(3^{rd}\) term is 12 and the last term is 106. Find the \(29^{th}\) term.
- If the \(3^{rd}\) and the \(9^{th}\) terms of an AP are 4 and –8 respectively, which term of this AP is zero?
- The \(17^{th}\) term of an AP exceeds its \(10^{th}\) term by 7. Find the common difference.
- Which term of the AP: 3, 15, 27, 39… will be 132 more than its \(54^{th}\) term?
- 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.
- How many three digit numbers are divisible by 7?
- How many multiples of 4 lie between 10 and 250?
- For what value of n, are the nth terms of two AP’s: 63, 65, 67… and 3, 10, 17… equal?
- Determine the AP whose third term is 16 and the \(7^{th}\) term exceeds the \(5^{th}\) term by 12.
- Find the \(20^{th}\) term from the last term of the AP: 3, 8, 13… , 253.
- 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.
- 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?
- 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.

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