Question

The first and the third term of an A.P., are equal, respectively, to the first and the third term of a G.P., and the second term of the A.P exceeds the second term of the G.P. by $0.25$. Calculate the sum of the first five terms of the A.P. if its first term is $2$.

• A. 22.5
• B. 2.5
• C. both A and B
• D. only 22.5

Answer 1

Answer: (C)

both A and B

Let the first three terms of an AP be $a,a+d,a+2d$ and GP be $A,Ar,A{r}^2$
given that $a=A;a+2d=A{r}^2$ and $a+d=Ar+0.25$
if $a=2$ given equations becomes
$1+d=r^2$ and $8+4d=8r+1$ -----(1)
eliminating $d$ from above equations gives
$4r^2-8r+3=0$
$\to 4r^2-8r+3=(2r-1)(2r-3)=0$
$\therefore$ $r=\frac{1}{2}orr=\frac{3}{2}$
when $r=\frac{1}{2}$
from (1) $d=-\frac{3}{4}$
when $r=\frac{3}{2}$
from (1) $d=\frac{5}{4}$
$\therefore$ sum of the first five terms of the A.P when $a=2$ and $d=-\frac{3}{4}$ is
$S=\frac{n}{2}(2a+(n-1\left)d\right)=\frac{5}{2}(4-3)=2.5$
sum of the first five terms of the A.P when $a=2$ and $d=\frac{5}{4}$ is
$S=\frac{n}{2}(2a+(n-1\left)d\right)=\frac{5}{2}(4+5)=22.5$
Hence, option C.