Question

The 10th,4th and 1st terms of an A.P are the three consecutive terms of the G.P and the sum of the 1st 6 terms, taking the 1st term to be 4.

Answer 1

10th term --- a + 9d
4th term --- a + 3d
1sterm ---- 4

so
(4+3d)/(4+9d) = 4/(4+3d)

(4+3d)^2 = 4(4+9d)
16 + 24d + 9d^2 = 16 + 36d
9d^2 - 12d = 0
d(9d - 12) = 0
d = 0 or d = 12/9

if d = 0, we could have terms in the AP that would not change, not very interesting, but nevertheless possible,
the terms would be 4, 4, 4, ....
and the sum of the first 6 terms would simply be 24
note that the 10th, the 4th, and the 1st would form a GP, since
4/4 = 4/4

if d = 12/9 = 4/3
sum(6) = 3(8 + 5(4/3) ) = 44

check: the terms of the AP would be
4, 16/3, 20/3, 8, 28/3, 32/3, 12, 40/3, 44/3, 16
is 8/16 = 4/8 ? YES
sum(6) = 4+16/3+ 20/3+ 8+28/3+32/3 = 44

My answer is correct