If the sum of the first 2 n terms of the A.P. 2 , 5 , 8 , … , is equal to the sum of the first n terms of the A.P. 57 , 59 , 61 , … , then n equals [2001]

Question

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots,$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots,$ then $n$ equals [2001]

Smart Achievers · Accepted Answer

(3)

$Given: 2 + 5 + 8 + \dots 2 n terms = 57 + 59 + 61 + \dots n terms$

$\Rightarrow \frac{2 n}{2} [4 + (2 n - 1) \cdot 3] = \frac{n}{2} [114 + (n - 1) \cdot 2]$

$\Rightarrow 6 n + 1 = n + 56$

$\Rightarrow 5 n = 55 \Rightarrow n = 11$

Solution

Q.
If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots,$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots,$ then $n$ equals [2001]

1 10

2 12

3 11

4 13

Ans.
(3)

$Given: 2 + 5 + 8 + \dots 2 n terms = 57 + 59 + 61 + \dots n terms$

$\Rightarrow \frac{2 n}{2} [4 + (2 n - 1) \cdot 3] = \frac{n}{2} [114 + (n - 1) \cdot 2]$

$\Rightarrow 6 n + 1 = n + 56$

$\Rightarrow 5 n = 55 \Rightarrow n = 11$

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