If $a_{5} = 12$ and $a_{15} = 400$ . Find $S_{10}$

314

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320

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316

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318

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Solution

The correct option is A 314
Using the formula,
$a_{n} = a + (n - 1) d$
So, $a_{5} = a + (5 - 1) d$
$12 = a + 4 d$ ............. (1)
$a_{15} = a + (15 - 1) d$
$400 = a + 14 d$ .......... (2)
Solving equation (1) & (2)
$12 = a + 4 d$
$400 = a + 14 d$
(-) (-) (-)
---------------------
$- 388 = - 10 d$
$d = 38.8$
Put the value of d in equation (1)
$12 = a + 4 (38.8)$
$12 = a + 155. 2$
$a = - 143.2$
The sum of first n term is given by $S_{n} = \frac{n}{2} (2 a + (n - 1) d)$
$S_{10} = \frac{10}{2} (2 (- 143. 2) + (10 - 1) 38.8)$
$= 5 (- 286.4 + 349.2)$
$= 5 (62.8)$
$S_{10} = 314$

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If a5=12 and a15=400. Find S10

If $a_{5} = 12$ and $a_{15} = 400$ . Find $S_{10}$