Question

In a certain $A.P.$ the ${24}^{th}$ term is twice the ${10}^{th}$ 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 10th term of an A.P is

$a_{10}=a_1+(10-1)d$

$a_{10}=a_1+9d\dots ...eq\left(2\right)$

and 24th term is $a_{24}$

$a_{24}=a_1+(24-1)d$

$a_{24}=a_1+23d$

given that the 24th term is twice the 10th term

$⟹a_{24}=2\times a_{10}$

$⟹2\times a_{10}=a_1+23d\dots ..eq\left(3\right)$

put value of $a_{10}$ from eq(2)

$⟹2\times (a_1+9d)=a_1+23d$

$⟹2a_1+18d=a_1+23d$

$⟹2a_1-a_1=23d-18d$

$⟹a_1=5d.....eq\left(4\right)$

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.

$a_{34}=5d+\left(33\right)d$

$a_{34}=38d\dots ...eq\left(5\right)$

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.

$a_{72}=5d+\left(71\right)d$

$a_{72}=76d$

$a_{72}=2\times 38d$

we know that from eq(5)

$a_{34}=38d$

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