Question

If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.

Answer 1

Given:

\[a_{m + 1} = 2 a_{n + 1} \]

\[ \Rightarrow a + (m + 1 - 1)d = 2[a + (n + 1 - 1)d]\]

\[ \Rightarrow a + md = 2(a + nd)\]

\[ \Rightarrow a + md = 2a + 2nd\]

\[ \Rightarrow 0 = a + 2nd - md \]

\[ \Rightarrow nd = \frac{md - a}{2} . . . (i)\]

To prove:

\[a_{3m + 1} = 2 a_{m + n + 1} \]

\[\text { LHS: } a_{3m + 1} = a + (3m + 1 - 1)d\]

\[ \Rightarrow a_{3m + 1} = a + 3md\]

\[RHS: 2 a_{m + n + 1} = 2[a + (m + n + 1 - 1)d]\]

\[ \Rightarrow 2 a_{m + n + 1} = 2(a + md + nd)\]

\[ \Rightarrow 2 a_{m + n + 1} = 2\left[ a + md + \left( \frac{md - a}{2} \right) \right] \left( \text { From }(i) \right)\]

\[ \Rightarrow 2 a_{m + n + 1} = 2\left[ \frac{2a + 2md + md - a}{2} \right]\]

\[\Rightarrow 2 a_{m + n + 1} = 2\left[ \frac{a + 3md}{2} \right]\]

\[ \Rightarrow 2 a_{m + n + 1} = a + 3md\]

∴ LHS = RHS