In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

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Solution

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

$To prove : a_{72} = 2 a_{34} LHS : a_{72} = a + (72 - 1) d \Rightarrow a_{72} = a + 71 d \Rightarrow a_{72} = 5 d + 71 d (From (i)) \Rightarrow a_{72} = 76 d$

$RHS : 2 a_{34} = 2 [a + (34 - 1) d] \Rightarrow 2 a_{34} = 2 (a + 33 d) \Rightarrow 2 a_{34} = 2 (5 d + 33 d) (Form (i)) \Rightarrow 2 a_{34} = 2 (38 d) \Rightarrow 2 a_{34} = 76 d$
∴ RHS = LHS
Hence, proved.

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