Question

If $(m+1)$th term of an A.P. is twice the $(n+1)$th term, prove that the $72$nd term is twice the $34$th term.

Answer 1

The $n$th term of an A.P is given by

$a_n=a_1+(n-1)d\dots ..eq\left(1\right)$

where $a_1$ is the first term of A.P

and d is the common difference

hence (m+1)th term of an A.P is

$a_{10}=a_{m+1}+(m+1-1)d$

$a_{m+1}=a_1+md\dots ...eq\left(2\right)$

and (n+1)th term is

$a_{n+1}=a_1+(n+1-1)d$

$a_{n+1}=a_1+nd$

given that the (m+1)th term is twice the (n+1)th term

$⟹a_{m+1}=2\times a_{n+1}$

$⟹2\times a_{n+1}=a_1+md\dots ..eq\left(3\right)$

put value of $a_{n+1}$ in eq(3)

$⟹2\times (a_1+nd)=a_1+md$

$⟹2a_1+2\times nd=a_1+md$

$⟹2a_1-a_1=(m-2n)d$

$⟹a_1=(m-2n)d.....eq\left(4\right)$

now 34th term of A.P is given by

$a_{34}=a_1+(34-1)d$

put value of $a_1$ from eq(4) in above equation.

$a_{34}=(m-2n)d+\left(33\right)d$

$a_{34}=(m-2n+33)d\dots ...eq\left(5\right)$

now 72nd term of A.P is given by

$a_{72}=a_1+(72-1)d$

put value of $a_1$ from eq(4) in above equation.

$a_{72}=(m-2n)d+\left(71\right)d$

$a_{72}=(m-2n+71)d$

if, $a_{72}=2\times a_{34}$