geometric sequence

1. 4 = a(r^10 - 1)/(r - 1) ................(1)
4 + 48 = 52 = a(r^30 - 1)/(r - 1) .................(2)
(2)/(1) we get
52/4 = (r^30 - 1)/(r^10 - 1) = 13
Let r^10 = x, we get
13 = (x^3 - 1)/(x - 1)
13(x - 1) = x^3 - 1
x^3 - 13x + 12 = 0
By Factor theorem, x = 1 is a factor, so the equation can be factorized as
(x - 1)(x - 3)(x + 4) = 0
so x = r^10 = 1, 3 or - 4.
r^10 = 1 and - 4 are rejected, because r cannot be equal to 1 and r must be real.
so only solution is r^10 = 3.......(4)
Sub. into (1), we get a/(r - 1) = 4/(r^10 - 1) = 4/(3 - 1) = 2.......(5)
Using the result of (4) and (5), Sum to 60th term = a(r^60 - 1)/(r - 1)
= [a/(r - 1)][(r^10)^6 - 1] = 2(3^6 - 1) = 2(729 - 1) = 1456
So sum of 31st term to 60th term = 1456 - 4 - 48 = 1404.

Let a be the first term and d be the common difference.

Sum of the first 10 terms:
T(1) + T(2) + T(3) + ...... + T(10) = 4
10[2a + (10 - 1)d]/2 = 4
10a + 45d = 4 ...... (1)

Sum from the 11th term to the 30 term:
T(11) + T(12) + T(13) + ...... T(30) = 48
[T(1) + T(2) + ...... + T(30)] - [T(1) + T(2) + ...... T(10)] = 48
30[2a + (30 - 1)d]/2 - 10[2a + (10 - 1)d]/2 = 48
30a + 435d - 10a - 45d = 48
20a + 390d = 48 ...... (2)

(2) - 2*(1) :
300d = 40
d = 2/15

Put d = 4/30 into (1):
10a + 45(2/15) = 4
10a + 6 = 4
10a = -2
a = -1/5

The sum from 31st to the 60th term
= T(31) + T(32) + ...... + T(60)
= [T(1) + T(2) + T(3) + ...... T(60)] - [T(1) + T(2) + T(3) + ...... + T(30)]
= {60[2*(-1/5) + (60 - 1)*(2/15)]/2} - 48
= {30[(-2/5) + (118/15)]} - 48
= 30*(112/15) - 48
= 224 - 48
= 176