First slide

Arithmetic progression

Question

Given sum of the first n terns of an AP is 2n + 3 n². Another AP is formed with the same first term and double of the common difference, the sum of n terms of the new AP is

Moderate

A

$n + 4 n^{2}$
B

$6 n^{2} - n$
C

$n^{2} + 4 n$
D

$3 n + 2 n^{2}$

Solution

Here, $T_{1} = S_{1} = 2 (1) + 3 (1)^{2} = 5$
$T_{2} = S_{2} - S_{1} = 16 - 5 = 11 [∵ S_{2} = 2 (2) + 3 (2)^{2} = 16]$
$T_{3} = S_{3} - S_{2} = 33 - 16 = 17$ $[∵ S_{3} = 2 (3) + 3 (3)^{2} = 33]$
Hence, sequence is 5, 11 , 17.
$∴ a = 5$ and $d = 6$
For new AP,
$\begin{array}{l} A = 5, D = 2 \times 6 = 12 \\ ∴ S_{n}^{'} = \frac{n}{2} [2 \times 5 + (n - 1) 12] \\ = 6 n^{2} - n \end{array}$

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