Q9 of 176 Page 8

Find the sum to first n terms of an A.P. whose kth term is 5k + 1.

As it is given that kth term of the AP = 5k + 1

∴ a_k = a + (k – 1)d

⇒ 5k + 1 = a + (k – 1)d

⇒ 5k + 1 = a + kd – d

Now, on comparing the coefficient of k, we get

d = 5

and a – d = 1

⇒ a – 5 = 1

⇒ a = 6

We know that,

Find the sum of first 25 terms of an A.P. whose nth term is 1 — 4n.

If the sum to n terms of a sequence be n² + 2n, then prove that the sequence is an A.P.

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

[Hint: t_m = S_m — S_m-1= 3m² + 5m — 3 (m— 1)² — 5 (m— 1) = 3 (2m — 1) + 5 = 6m + 2]

If the sum of n terms of an A.P. is pn + qn², where p and q are constants, find the common difference.