Given an A.P.

Question

Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater than 200 and less than 220. If the second term in it is 12, then its 4th term is:

Accepted Answer

Let $a$ = first term, $d$ = common difference: $a+d=12$ ... (i)

$S_9=\dfrac92(2a+8d)=9(a+4d)=9(a+d+3d)=9(12+3d)$.

$$200<9(12+3d)<220 \Rightarrow 200<108+27d<220 \Rightarrow 92<27d<112.$$

The only integer value giving positive-integer terms is $d=4$ (since $27\times4=108$, and $92<108<112$).

Then $a=12-4=8$, so $4$th term $=a+3d=8+12=20$.

Solutions · 1