Find the sum of first 31 terms of an A.P. whose third term is 12 and fourth term is 16.

1,983

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1,984

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1,985

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1,986

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Solution

The correct option is B 1,984
Given that, $a_{3} = 12; a_{4} = 16$
Common difference, $d = a_{4} - a_{3} = 16 - 12 = 4$
$a_{3} - a_{2} = d$
$12 - 4 = a_{2}$ $\Rightarrow 8$
$d = a_{2} - a_{1}$
$a = 4$
We know the formula,
$s_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{31} = \frac{31}{2} [2 \times 4 + (31 - 1) 4]$
$= 15.5 [8 + 30 \times 4]$
$= 15.5 [128]$
$S_{31} = 1, 984$

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