The sum of first m terms of an AP is (4m2 - m). If its nth term is 107, find the value of n. Also, find the 21st term of this AP.
Let a be the first term and d be the common difference.
Given: Sum of first m terms of an AP is given by:
Sm = [2a + (m - 1)d] = 4m2 - m
Now, nth term is given by: an = Sn - Sn - 1
∴ an = (4n2 - n) - [4(n - 1)2 - (n - 1)]
= (4n2 - n) - [4(n2 + 1 - 2n) - n + 1]
= 4n2 - n - 4n2 - 4 + 8n + n - 1
= 8n - 5 …………………(1)
Now, given that an = 107
⇒ 8n - 5 = 107
⇒ 8n = 112
⇒ n = 14
For 21st term of AP, put n = 21 in the value of the nth term in equation (1), we get
a21 = 8 × (21) - 5
⇒ a21 = 168 - 5
= 163
∴ a21 = 163