Question

Find four numbers in gp such that the sum of the extreme numbers s 112 and the sum of the middle terms is 48

Answer 1

Let the extreme terms be a and d
and the middle terms be b and c

Thus According to question

a + d = 112
b + c = 48

Thus    Total Sum = 112 +  48  = 160

Thus     S(4)  = a ( 1 - r^n) / (1 - r)
             160  =  a (1 - r^4) / (1 - r)
              160 = a (1+r) (1+r^2)

Also      S(2 ) = b (1-r^2) / (1-r)
              48 = b ( 1+r )
also  b/a = r
So       b = a r
Thus         48 = ar ( 1+r )
                 48 = r * 160 / (1+r^2)
                  48 +  48 r^2 = 160 r
                 48 r^2 - 160 r + 48 = 0
                  12 r^2 - 40 r + 12 = 0
                   3 r^2 - 10r + 3 = 0
                    3 r^2 - 9r - r + 3 = 0
                    3r (r - 3) - (r - 3) = 0
                         r = 3 or 1/3
So    b/a = 1/3
            b = a/3