The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term. [CBSE 2013]

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Solution

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

$a_{24} = 2 a_{10}$ (Given)
$\Rightarrow a + 23 d = 2 (a + 9 d) [a_{n} = a + (n - 1) d] \Rightarrow a + 23 d = 2 a + 18 d \Rightarrow 2 a - a = 23 d - 18 d \Rightarrow a = 5 d . . . . . (1)$

Now,
$\frac{a_{72}}{a_{15}} = \frac{a + 71 d}{a + 14 d} \Rightarrow \frac{a_{72}}{a_{15}} = \frac{5 d + 71 d}{5 d + 14 d} [From (1)] \Rightarrow \frac{a_{72}}{a_{15}} = \frac{76 d}{19 d} = 4 \Rightarrow a_{72} = 4 \times a_{15}$

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

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

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