Question

Let $a,b,c$ be in G.P with common ratio, where $a\ne 0$ and $0<r\le \frac{1}{2}$. If $3a,7b,15c$ are the first three terms of an A.P., then the 4th term of thisA.P is

• A. $\frac{7}{3}a$
• B. $a$
• C. $\frac{2}{3}a$
• D. $5a$

Answer 1

Answer: (B)

$a$

$b=ar;c=a{r}^2$

$\Rightarrow 14b=3a+15c\Rightarrow 14\left(ar\right)=3a+15a{r}^2\Rightarrow 14r=3+15r^2$

$\Rightarrow 15r^2-14r+3=0\Rightarrow (3r-1)(5r-3)=0\Rightarrow r=\frac{1}{3},\frac{3}{5}$
Only acceptable value is $r=\frac{1}{3}$, because
$r\in (0,\frac{1}{2}]$
$\therefore d=7b-3a=7r-3a=\frac{7}{3}a-3a=-\frac{2}{3}a$
4th term $=15c-\frac{2}{3}a=\frac{15}{9}a-\frac{2}{3}a=a$