# 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.

Let a be the first term and d be the common difference.

Given: a10 = 41

a16 = 5 a3

Now, Consider a10 = 41

a + (10 - 1)d = 41

a + 9d = 41 ………………….(1)

Consider a16 = 5 a3

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

a + 15d = 5a + 10d

4a - 5d = 0 ………………….(2)

Now, subtracting equation (2) from 4 times of equation (1), we get,

41d = 164

d = 4

from equation (2), we get,

4a= 5d

4a = 20

a = 5

Now, Sum of first n terms of an AP is

Sn = [2a + (n - 1)d]

Sum of first 15 terms is given by:

S15 = [2(5) + (15 - 1)(4)]

= (15/2) × [10 + 56]

= 15 × 33

= 495

S15 = 495

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