Question

The 24^th term of an A.P. is twice its 10^th term. Show that its 72^nd term is 4 times its 15^thterm.

Answer 1

Let a be the first term and d be the common difference.
We know that, n^th term = a_n = a + (n − 1)d
According to the question,
a₂₄ = 2a₁₀
⇒ a + (24 − 1)d = 2(a + (10 − 1)d)
⇒ a + 23d = 2a + 18d
⇒ 23d − 18d = 2a − a
⇒ 5d = a
⇒ a = 5d   .... (1)
Also,
a₇₂ = a + (72 − 1)d
      = 5d + 71d              [From (1)]
      = 76d                      ..... (2)
and
a₁₅ = a + (15 − 1)d
      = 5d + 14d              [From (1)]
      = 19d                      ..... (3)
On comparing (2) and (3), we get
76d = 4 × 19d
⇒ a₇₂ = 4 × a₁₅
Thus, 72^nd term of the given A.P. is 4 times its 15^th term.