Question

If $10$ times the ${10}^{th}$ term of an $A.P.$ is equal to $15$ times the ${15}^{th}$ term, show that $25th$ term of the $A.P.$ is zero.

Answer 1

We know that the $n^{th}$ term of the arithmetic progression is given by $a+(n-1)d$

Given that the 10 times the ${10}^{th}$ term is equal to $15$ times the ${15}^{th}$ term

Therefore, $10\left({10}^{th}term\right)=15\left({15}^{th}term\right)$

$⟹10(a+(10-1\left)d\right)=15(a+(15-1\left)d\right)$

$⟹10(a+9d)=15(a+14d)$

$⟹10a+90d=15a+210d$

$⟹15a-10a=90d-210d$

$⟹5a=-120d$

$⟹a=-24d$ ------(1)

The ${25}^{th}$ term is $a+(25-1)d=a+24d=-24d+24d=0$

Therefore, the ${25}^{th}$ term of the A.P. is zero