If the sum of the first 20 terms of an A.P. is 1050 and its first term is 5, find the sum of 12 terms.

190

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290

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390

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490

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Solution

The correct option is C $390$
First term $= 5$
$S_{20} = 1050$
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{20} = \frac{20}{2} [2 \times 5 + (20 - 1) d]$
$1050 = 10 [10 + 19 d]$
$\frac{1050}{10} = 10 + 19 d$
$105 - 10 = 19 d$
$95 = 19 d$
$d = 5$
So, sum of 12th term is
$S_{12} = \frac{12}{2} [2 \times 5 + (12 - 1) 5]$
$S_{12} = 6 [10 + 55]$
$S_{12} = 390$

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