Question

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

Answer 1

Let our number's digits be ar² ar a
So, we have our number as
ar²(100) + ar(10) + a

Now, if we subtract 792 from this, we get our digits reversed, so
ar²(100) + ar(10) + a - 792 = a(100) + ar(10) + ar²

100ar² + 10ar + a - 792 = 100a + 10ar + ar²
99ar² - 792 = 99a
ar² - 8 = a
a(r² - 1) = 8....(1)
Now if we increase tens place by 2, we get digits in AP, so
ar², ar + 2, a are in AP
2(ar + 2) = ar² + a
2ar + 4 = ar² + a
ar² - 2ar + a = 4
a(r² - 2r + 1) = 4
a(r - 1)² = 4.....(2)
Solving (1) and (2), we get
a(r - 1)(r + 1)/a(r - 1)² = 8/4
r + 1/ r - 1 = 2
r + 1 = 2r - 2
r = 3
So, a = 1
So our digits are 9, 3, 1
Our number is 931

Answer 2

ANSWER : 931

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