Question

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

Answer 1

Given:

\[a_{24} = 2 a_{10} \]

\[ \Rightarrow a + \left( 24 - 1 \right)d = 2\left[ a + \left( 10 - 1 \right)d \right]\]

\[ \Rightarrow a + 23d = 2(a + 9d)\]

\[ \Rightarrow a + 23d = 2a + 18d\]

\[ \Rightarrow 5d = a . . . (i)\]

\[\text { To prove }: \]

\[ a_{72} = 2 a_{34} \]

\[\text { LHS: } a_{72} = a + \left( 72 - 1 \right)d\]

\[ \Rightarrow a_{72} = a + 71d\]

\[ \Rightarrow a_{72} = 5d + 71d \left( \text { From }(i) \right)\]

\[ \Rightarrow a_{72} = 76d\]

\[\text { RHS }: 2 a_{34} = 2\left[ a + \left( 34 - 1 \right)d \right]\]

\[ \Rightarrow 2 a_{34} = 2\left( a + 33d \right)\]

\[ \Rightarrow 2 a_{34} = 2(5d + 33d) \left( \text { Form }(i) \right)\]

\[ \Rightarrow 2 a_{34} = 2\left( 38d \right)\]

\[ \Rightarrow 2 a_{34} = 76d\]

∴ RHS = LHS
Hence, proved.