Question

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

Answer 1

$T_9$ of an AP = 6 $T_2$

$a+(9-1)d=6(a+(2-1)d$

$a+8d=6(a+d)$

$a+8d=6a+6d$

$8d-6d=6a-a$

$2d=5a$

$d=\frac{5}{2}a$

Now its given that $5$th term is $22$

That is $T_5$ = 22

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

$22=a+4d$

$22=a+4$ $\frac{5}{2}$ $a$ (the value which we found above for d )

$22=11a$

$a=2$

$d=\frac{5}{2}a$

Substituting the value of $a=2$ we have

$d=\frac{5}{2}\times 2$

$d=5$

Now the AP is: $2,7.12,17,22,27.$