math
posted by Edwin on .
(b) the digits of a positive integer having three digits are in A.P and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
(c) if a,b,c are in A.P, prove that (i) (ab)^1, 1/ac, 1/bc are also in A.P. (ii) a(b+c)/bc, b(a+c)/ca, c(a+b)/ab are also in A.P.
(d) if a^2, b^2, c^2 are in A.P, prove that (i) 1/(b+c), 1/(c+a), 1/(a+b) are also in A.P. (ii) a/(b+c), b/(c+a), c/(a+b) are also in A.P.
(e) the sum to infinity of a G.P series is R. The sum to infinity of the squares of the terms is 2R. The sum to infinity of the cubes of the terms is (64/13)R. Find (i) the value of R. (ii) the first term of the first original series.

(b) the digits of a positive integer having three digits are in A.P and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
If the three digits are in A.P. (arithmetic progression) and the sum is 15, the middle digit must be 5.
The only choices are 951, 852, 753, 654. Note: the original number is greater than the number with digits reversed.
Check which one fits the bill.
Give a try for the remaining problems and we'll be glad to provide help. 
1.
first equation a + a+d + a+2d = 15 or a+d=5
second equation
(100a + 10(a+d) + a+2d  (100(a+2d) + 10(a+d) + a = 594
solve ,
I got d=3 and a=8 , which works