Question

Find the three digit number whose consecutive number form a G.P. If we subtract 792 from this number, we get a number consisting of the same digits written in the reverse order. Now if we increase the second digit of the required number by 2, the resulting number will form an A.P.

• A. 901
• B. 931
• C. 981
• D. 991

Answer 1

Answer: (B)

931

Answer 2

Let the three digit be a, ar, ar² then according to hypothesis

100a + 10ar + ar² +792 = 100ar² + 10ar + a

⇒ $a\left( r^{2} - 1 \right) = 8$ .....................................(1)

and a, ar + 2, ar² are in A.P.

then $2(ar + 2) = a + ar^{2}$

⇒$a\left( r^{2} - 2r + 1 \right) = 4$.........................(2)Dividing (1) by (2),

Then $\frac{a\left( r^{2} - 1 \right)}{a\left( r^{2} - 2r + 1 \right)} = \frac{8}{4}$

⇒ $\frac{(r + 1)(r - 1)}{(r - 1)^{2}} = 2$⇒ $\frac{r + 1}{r - 1} = 2$

∴ r = 3 from (1), a =1

Thus digits are 1, 3, 9 and so the required number is 931.