Question

Sum of $4$ numbers in GP is $60$. And the AM of first and last no. is $18$ find the first term and common difference of the GP

• A. $a=4,r=2$
• B. $a=32,r=\frac{1}{2}$
• C. $a=3,r=1$
• D. $a=6,r=3$

Answer 1

Answer: (A), (B)

$a=32,r=\frac{1}{2}$
$a=4,r=2$

Let 4 numbers in G.P. be $a,ar,{ar}^2{,ar}^{3}a(1+r+r^2+r^3)=60\frac{a+{ar}^3}{2}=18=>a+{ar}^3=36a(1+r)(1+r^2)=60a(1+r)(1+r^2-r)=36\frac{1+r^2-r}{1+r^2}=\frac{36}{60}5+5r^2-5r=3+3r^{2}2r^2-5r+2=0r=2,\frac{1}{2}ifr=2,a(1+8)=36=>a=4ifr=\frac{1}{2},a(1+\frac{1}{8})=36=>a=32a=4,r=2\left(or\right)a=32,r=\frac{1}{2}$