Question

Three positive numbers form a GP. If the middle number is increased by $8$ , the three numbers form an AP. If the last number is also increased by $64$ along with the previous increase in the middle number, the resulting numbers form a GP again. Then

• A. common ratio $=3$
• B. first number $=4/9$
• C. common ratio $=-5$
• D. first number $=5$

Answer 1

Answer: (A)

Let three positive numbers which are in G.P. be $a,b,c$
$\therefore b^2=ac\dots .\text{ (i) }$
By first condition $a,b+8,c$ are in A.P.
$\therefore b=\frac{a+c}{2}-8$......(ii)
By second condition, $a,b+8,c+64$ are in G.P.
$\therefore (b+8{)}^2=a(c+64)$
$\therefore b^2+64+16b=ac+64a$
$\therefore b^2=ac$
$\therefore 64+16b=64a$
$\Rightarrow b=4(a-1)$........(iii)
putting the value of $b$ in (ii)
$4(a-1)=\frac{a+c}{2}-8$
$\Rightarrow c=7a+8$ ....(iv)
Now putting the value of both $b$ and $c$ in (i)
$16(a-1{)}^2=a(7a+8)$
$\Rightarrow 9a^2-4a+16=0$
$(9a-4)(a-4)=0$
$\Rightarrow a=4/9,a=4$
$a=4/9$ gives negative value of $b$, but it is given that $b$ is positive, so $a=4$ is acceptable value $b=4(4-3)=12$
$\therefore$ common ratio $=\frac{12}{4}=3$