Question

The sum of three numbers in G.P. is 14, If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers.

Answer 1

Let the numbers are : $\frac{a}{r}$, a and ar

Then

$\frac{a}{r}+a+ar=14$

Again the numbers a + 1, ar + 1 and (ar^2 - 1) are in A.P., therefore

$2(a+1)=(ar-1)+(\frac{a}{r}+1)$

$2(a+1)=ar+\frac{a}{r}$

$2(a+1)=14-a$

$3a=12$

$a=4$

Now we have

$\Rightarrow \frac{4}{r}+4+4r=14$

$\Rightarrow 2-5r+2r^2=0$

$\Rightarrow 2r^2-4r-r+2=0$

$\Rightarrow 2r(r-2)-1(r-2)=0$

$\Rightarrow (r-2)(2r-1)=0$

$\Rightarrow r=2,\frac{1}{2}$

Hence, the G.P. for a = 4 and

r = 2 is 8, 4, 2

And the G.P. for a = 4 and

$r=\frac{1}{2}is2,4,8,......$