Question

The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.

Answer 1

Let a be the first term and d be the common difference of the AP. Then,

a₂₄ = 2a₁₀   ...(Given)

⇒ a + 23d = 2(a + 9d)   ...[a_n = a + (n – 1)d]

⇒ a + 23d = 2a – 18d

⇒ 2a – a = 23d – 18d

⇒ a = 5d   ...(1)

Now, 

`(a_72)/(a_15) = (a + 71d)/(a + 14d)`

⇒ `(a_72)/(a_15) = (5d + 71d)/(5d + 14d)`   ...[From (1)]

⇒ `(a_72)/(a_15) = (76d)/(19d) = 4`

⇒ `a_72 = 4 xx a_15`

Hence, the 72^nd term of the AP is 4 times its 15^th term.