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

$t_{19}=3t_6$

$\Rightarrow 3(a+(6-1)d)=a+(19-1)d$

$3(a+5d)=a+18d$

$3a+15d=a+18d$

$2a=3d$

$\Rightarrow d=\frac{2}{3}a$

$\Rightarrow t_9=19$

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

$a+8d=19$

$a+8\times \frac{2}{3}a=19$

$3a+16a=19\times 3$

$19a=19\times 3$

$\Rightarrow a=3$

$\Rightarrow d=\frac{2}{3}a=\frac{2}{3}\times 3=2$

$\therefore$ A.P is $3,5,7,\mathrm{9......}$

Hence, solved.