The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16, find the sum of its first 10 terms.

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Solution

$a_{13} = 4 \times a_{3}$

$a_{5} = 16$

as we know $a_{n} = a + (n - 1) d$

so $a_{13} = a + 12 d$

$a_{3} = a + 2 d$

equating

$a + 12 d = 4 a + 8 d$

we get,

$a = \frac{4}{3} \times d$

also, $a_{5} = a + 4 d = 16$

putting value of a we get,

d=3

also a=4

as we know,

$S_{n} = \frac{n}{2} \times (a_{1} + a_{n})$

$a_{10} = 4 + (9 \times 3) = 31$

$S_{10} = 5 \times (4 + 31) = 175$

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