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$=\frac{4}{9}$
• C. common ratio$=-5$
• D. first number $=4$

Answer 1

Answer: (A), (D)

The correct options are
A common ratio$=3$
D first number $=4$

Let the three numbers be $\frac{a}{r},aandar$

Now A/Q $2(a+8)=(\frac{a}{r}+ar)$

Also A/Q ${(a+8)}^2=\frac{a}{r}\times (ar+64)$

Solving the above two equations :

From the 2nd equation $a^2+64+16a=a^2+\frac{64a}{r}\Rightarrow 4+a=\frac{4a}{r}\Rightarrow a=\frac{4r}{4-r}$

Substituting this in 1st equation

$2(\frac{4r}{4-r}+8)=(\frac{4}{4-r}+\frac{4r^2}{4-r})$

$\Rightarrow 16(4-r)=4r^2-8r+4\Rightarrow r^2-2r+1=16-4r\Rightarrow r^2+2r-15=0\Rightarrow r=3,-5anda=\frac{4r}{4-r}=12or\frac{-20}{9}$

but the second term is neglected because according to question the numbers are positive.

Hence common ratio$=3$ and

first term $=\frac{a}{r}=\frac{12}{3}=4$