Question

Digits of a three digit number are in g.p. if one is added to the middle digit, the digits form an
a.p. if the digits of the unit place and the hundred's place are interchanged, then the new number is 594 more than the original number. the digit at unit's place of original number, is

Answer 1

Let the three digits in order of hundreds, tens and ones be....... a/r, a, ar where r is the common ratio of the gp

Thus the number will be
100*a/r + 10a + ar

Now given that, a/r, a+1, ar are in ap

We know that when a, b, c are in ap then
2b = a+c

Therefore,

2(a+1) = a/r + ar
2a +2 = a(r + 1/r)
a(r + 1/r - 2) = 2......(1)


Also given that,

100ar + 10a + a/r = 100*a/r + 10a + ar + 594
=> 100ar + a/r = 100a/r + ar +594
=> 99 ar = 99a/r + 594
=> ar = a/r + 6
=> a(r - 1/r) = 6.......(2)

Comparing equations (1) and (2),
Eqn (2) = 3* eqn (1)

a(r - 1/r) = 3( a (r + 1/r - 2))
r - 1/r = 3r + 3/r - 6
2r + 4/r = 6
=> r + 2/r = 3
=> r^2 - 3r +2 = 0
=>
By solving this quadratic equation, you get r = 1 and r = 2.
We reject r = 1 as if we put r=1 in equation (2), a will be undefined.

Hence r = 2 is valid. When we substitute r=2 in equation (2), we get,

a(2 - 1/2) = 6
a = 6/1.5
a = 4

Hence the digits are
At hundreds place a/r = 4/2 =2
At tens place a = 4
At ones place ar = 4*2 = 8

Therefore the original number is 248 and digit at unit's place is 8