Question

The integers $2,a,b,c,160$ are such that the first three terms in $A.P.$ while the last three terms are in $G.P.$ If the common ratio of the $G.P.$ and common different of the $A.P.$ are equal, then the value of $a+b+c$ is ?

• A. $56$
• B. $64$
• C. $72$
• D. $Noneofthese$

Answer 1

The correct option is A $56$

$\underset{Ap}{\underset{}{2,a,b,}}\underset{Gp}{\underset{}{c,160}}$ Consider 'd' in Common difference

Consider 'r' in common ratio

$2a=6+2$ for Ap

$c^2=160b$ for Gp

$a-2=b-a=dr=\frac{c}{b}=\frac{160}{c}$

$As,d=r,b-a=\frac{c}{b}=a-2\Rightarrow b=2a-2$

$b-a=\frac{c}{b}$ ...(1)

$2a-2-a=\frac{c}{b}$

$a-2=\frac{160}{c}$ ...(2)

$c=\frac{160}{a-2}$

$a-2=\frac{c}{2(a-1)}$

$c=2(a-1)(a-2)$ ...(3)

Equating $\left(3\right)and\left(2\right)$

$\frac{160}{a-2}=2(a-1)(a-2)\Rightarrow 160=2(a-1)(a-2{)}^2$

By trial & error , we get $a=6.$

Substitute a in (3)

$\Rightarrow c=2\left(5\right)\left(4\right)=40$

$2a=b+2\Rightarrow 12-2=6=10$

$a+b+c=6+10+4=56$