Q. 334.1( 9 Votes )

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 :

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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