Question

The $19$th term of an A.P. is equal to three times its sixth term. If its $9$th term is $19$, 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 ${19}^{th}$ term is three times the $6^{th}$ term.

Therefore, ${19}^{th}term=3\left(6^{th}term\right)$

$⟹a+(19-1)d=3(a+(6-1\left)d\right)$

$⟹a+18d=3a+15d$

$⟹2a-3d=0$ -------(1)

Given that the $9^{th}$ term is $19$

Therefore, $a+(9-1)d=19$

$⟹a+8d=19$ ------(2)

2*eqn (2)- eqn (1) we get

$2(a+8d)-(2a-3d)=19$

$⟹19d=38$

$⟹d=2$

substituting $d=2$ in (1) we get

$2a-3\left(2\right)=0$

$⟹2a=6$

$⟹a=2$

Therefore, $$3, 5, 7, 9.....$ is the required sequence.