Question

The sum of infinite A.G.P. $3,4,4,\frac{32}{9}$............is

• A. $16$
• B. $18$
• C. $24$
• D. $27$

Answer 1

Answer: (D)

$27$

$a+(a+d)r+(a+2d)r^2+...........\infty$

$a=3,(3+d)r=(3+2d)r^2$ =4

Solving the last two, we have

$\frac{4}{r}-3=(\frac{4}{r^2}-3).\frac{1}{2}$

or$(4-3r).2r=4-3r^2$

or $3r^2-8r+4=0$ or$(r-2)(3r-2)=0$

$\therefore r=\frac{2}{3}$ as $r<1$

$\therefore d=3$

$S=3+6.\frac{2}{3}+9(\frac{2}{3}{)}^2+..........\infty$

$\frac{2}{3}S=3.\frac{2}{3}+6(\frac{2}{3}{)}^2+...........\infty$

$\therefore S(1-\frac{2}{3})=3+3.\frac{2}{3}+3.(\frac{2}{3}{)}^2+..........\infty$

$\frac{S}{3}=\frac{3}{1-\left(2/3\right)}=9$

$\therefore S=27$