Question

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

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

Answer 1

The correct option is B $a$

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

$3a,7b,15c\to A.P.$
$14b=3a+15c$
$14\left(ar\right)=3a+15\left(a{r}^2\right)$
$15r^2-14r+3=0$
$\Rightarrow r=\frac{1}{3},\frac{3}{5}\text{(rejected)}(\because 0<r\le \frac{1}{2})$

Common difference $=7b-3a$
$=7ar-3a$
$=\frac{7a}{3}-3a$
$=-\frac{2}{3}a$
$4^{th}\text{term is}\Rightarrow 15c-\frac{2}{3}a=\frac{15}{9}a-\frac{2}{3}a=a$