Question

The $9$th term of an A.P is equal to $6$ times its second term. If its $5$th term is $22$, find the A.P.

Answer 1

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

Given that the $9^{th}$ term is six times the $2^{nd}$ term.

Therefore, $9^{th}term=6\left(2^{nd}term\right)$

$⟹a+(9-1)d=6(a+(2-1\left)d\right)$

$⟹a+8d=6a+6d$

$⟹5a-2d=0$ -------(1)

Given that the $5^{th}$ term is $22$

Therefore, $a+(5-1)d=22$

$⟹a+4d=22$ ------(2)

On solving we get

$⟹d=5$

$a=2$

Therefore, $2,5,7,12,\mathrm{17.....}$ is the required sequence.