First slide

Arithmetic progression

Question

The sum of first three terms of an increasing A.P. is $9$ and sum of their squares is $35 .$ The sum to $n$ terms of the series is

Moderate

A

$3 n^{2}$
B

$2 n^{2}$
C

$6 n - 2 n^{2}$
D

$n^{2}$

Solution

Let $d$ be common difference of the A.P., then
$(a_{2} - d) + a_{2} + (a_{2} + d) = 9$
and ${(a_{2} - d)}^{2} + a_{2}^{2} + {(a_{2} + d)}^{2} = 35$
$\Rightarrow a_{2} = 3$ and $3 a_{2}^{2} + 2 d^{2} = 35$
$\Rightarrow d = 2 .$
$\begin{array}{l} S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d] = \frac{n}{2} [2 a_{2} + (n - 3) d] \\ = \frac{n}{2} [2 (3) + (n - 3) (2)] = n^{2} \end{array}$

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