Question

The sum of first ten terms of an A.P. is 155 and the sum of first two terms of a G.P is 9. The first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to the common difference of the A.P which can be the A.P. as per the given conditions?

• A. 2, 4, 6, 8, 10...
• B. $\frac{25}{2},\frac{79}{6},\frac{83}{6},...$
• C. 2, 5, 8, 11,...
• D. both (b) and (c)

Answer 1

The correct option is D

both (b) and (c)

When you check option (a) it will be proved wrong. Again for convenience consider option (c)
2, 5, 8, 11...
then the first term of that G.P. = 3
and the common ratio of G.P = 2
Hence G.P. = 3, 6, 12, 24,...
So $a_1+a_2=9$ which is correct
and $S_n$ of A.P. = $\frac{10}{2}[2\times 2+9\times 3]=155$
which is also correct. Hence choice (c) is correct.
Similarly choice (b) is also correct.
Hence choice (d) is most appropriate.

Alternatively:
a, a + d, a + 2d, ... be an A.P.
and A. Ar. $A{r}^2,A{r}^3$... be a G.P.
then $\frac{n}{r}[2a+(n-1\left)d\right]=155$
2a + 9d = 31..........(1)
Again A + Ar = 9
d + ad = 9.......(2)
Solving equations (1) and (2) by substitution method,
$a=2,\frac{25}{2}andd=3,\frac{2}{3}$
Thus $A.P.\to 2,5,8,1,...G.P.\to 3,6,12,24,...orA.P.\to \frac{25}{2},\frac{79}{6},\frac{83}{6},...G.P.\to \frac{2}{3},\frac{25}{3},\frac{625}{6},...$
Thus choice (d) is correct.