Question

The 16th term of an AP is 5 times its 3rd term. If its 10th term is 41, find the sum of its first 15 terms.

Answer 1

16th term = a + 15d
3rd term = a + 2d

given

5(a + 2d) = a + 15d
5a + 10d = a + 15d

4a - 5d = 0

4a = 5d

$\Rightarrow a=\frac{5d}{4}$

Also,

10th term = a + 9d = 41

Substitute value of a here,

so,
$\frac{5d}{4}+9d=41$

$\frac{41d}{4}=41$

41d = 164

d = 4
a + 9d = 41
a + 36 =41
a = 5

n = 15
a = 5
d = 4

Sum of n terms, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$=7.5(10+14\times 4)$
$=7.5\times 66=495$