Q24 of 102 Page 410

If the sum of n terms of an AP is (3n2 + 5n) and its mth term is 164, find the value of m.

To Find: m


Given: Sum of n terms, mth term


Put n = 1 to get the first term


So a1 = 3 + 5 = 8


Put n = 2 to get the sum of first and second term


So a1 + a2 = 12 + 10 = 22


So a2 = 14


Common difference = 14 - 8 = 6


Tn = a + (n - 1)d = 8 + (n - 1)6 = 6n + 2


Now 6m + 2 = 164


Or m = 27


The value of m is 27.


More from this chapter

All 102 →
22

Find the sum of all integers between 100 and 600, each of which when divided by 5 leaves 2 as remainder.

 23

The sum of first 7 terms of an AP is 10 and that of next 7 terms is 17. Find the AP.

 25

Find the sum of all natural numbers from 1 and 100 which are divisible by 4 or 5.

 26

If the sum of n terms of an AP is , where P and Q are constants then find the common difference.