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

Let the first term of A.P. be a

and the common difference be d

$T_{m+1}=a+(m+1-1)d$

$=a+md$

$T_{n+1}=a+nd$

Given: $T_{m+1}=2T_{n+1}$

$\Rightarrow a+md=2(a+nd)$

$a=d(m-2n)$

$a=md-2nd\Rightarrow nd=\frac{md-a}{2}\to 1$

$T_{3m+1}=a+3md\to 3$

$T_{m+n+1}=a+(m+n)d$

$=a+md+nd$

Putting $1$ in $T_{m+n+1}$

$\Rightarrow T_{m+n+1}=a+md+\frac{md-a}{2}$

Multiplying both sides by $2$

$\Rightarrow 2T_{m+n+1}=2a+2md+md-a$

$=a+3md=3$

Hence, $2T_{m+n+1}=T_{3m+1}$