Suppose that the number of terms in an A.P. is 2k, . If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : [2025]

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Solution

Q.
Suppose that the number of terms in an A.P. is 2k, $k \in N$ . If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : [2025]

1 6

2 4

3 5

4 8

Ans.
(3)

Let the A.P. be $a_{1}, a_{2}, a_{3} . . . a_{2 k}$

Given, $\sum_{r = 1}^{k} a_{2 r - 1} = 40, \sum_{r = 1}^{k} a_{2 r} = 55 and a_{2 k} - a_{1} = 27$

$\Rightarrow \frac{k}{2} [2 a_{1} + (k - 1) 2 d] = 40$ ... (i)

and $\frac{k}{2} [2 a_{2} + (k - 1) 2 d] = 55$ ... (ii)

and $a_{1} + (2 k - 1) d - a_{1} = 27$ ... (iii)

After solving, we get

From (i), $a_{1} = \frac{40}{k} - (k - 1) d$

From (ii), $a_{2} = \frac{55}{k} - (k - 1) d$

and from (iii), $d = \frac{27}{2 k - 1}$

(ii) –(i)

$\Rightarrow d = \frac{15}{k} \Rightarrow \frac{27}{2 k - 1} = \frac{15}{k} \Rightarrow 9 k = 10 k - 5$

$∴$ k = 5.

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