A GEOMETRIC series question! help

the sum of the first 5 terms of a geometric sequence is 77 and the sum of the next 5 terms is -2464. find the 4th term of the sequence.

The sum of n terms of a geometric series with starting value "a" and common ratio "r" is
.. sum = a(r^n-1)/(r-1)

We are given two series sums: for 5 terms and for 10 terms.
.. sum_5 = a(r^5-1)/(r-1) = 77
.. sum_10 = a(r^10-1)/(r-1) = -2464+77 ... the sum of the first 10 terms is the sum of the two 5-term series sums

Taking the ratio of these sums, we have
.. sum_10/sum_5 = -2387/77 = -31 = (r^10-1)/(r^5-1)
The numerator of this expression can be factored, so the expression can be simplified.
.. -31 = (r^5+1)(r^5-1)/(r^5-1) = r^5+1 ... factor the difference of two squares (r^10 = (r^5)^2)
.. -32 =r^5
.. (-2)^5 = r^5
.. -2 = r

We can find a from
.. sum_5 = 77 = a(-32-1)/(-2-1) = 11a
.. 7 = a

So, the 4th term of the series is
.. 4th term = ar^3 = 7*(-2)^3 = -56

Hello,

Let uₐ be your geometric sequence with q as reason.
We are told:
u₀ + u₁ + u₂ + u₃ + u₄ = 77
u₅ + u₆ + u₇ + u₈ + u₉ = -2464

Since in any geometric sequence, we have:
uₐ = u₀.rª

77 = u₀ + u₁ + u₂ + u₃ + u₄
   = u₀ + u₀.r + u₀.r² + u₀.r³ + u₀.r⁴
   = u₀(1 + r + r² + r³ + r⁴)
   = u₀(r⁵ - 1) / (r - 1)

-2464 = u₅ + u₆ + u₇ + u₈ + u₉
   = u₀.r⁵ + u₀.r⁶ + u₀.r⁷ + u₀.r⁸ + u₀.r⁹
   = u₀.r⁵(1 + r + r² + r³ + r⁴)
   = u₀.r⁵(r⁵ - 1) / (r - 1)
   = [u₀(r⁵ - 1) / (r - 1)] × r⁵

Since u₀(r⁵-1)/(r-1)=77 :
-2464 = 77 × r⁵
r⁵ = -2464/77 = -32 = (-2)⁵
r = -2

Then, since r=-2 :
u₀(r⁵ - 1) / (r - 1) = 77
u₀(-32 - 1) / (-2 - 1) = 77
u₀(-33 / -3) = 77
u₀×11 = 77
u₀ = 7

And lastly, since uₐ=u₀.rª, the fourth term of the sequence is u₃, and its value is:
u₃ = u₀.r³ = 7.(-2)³ = 7 × -8 = -56

Methodically,
Dragon.Jade :-)

a(1+r+r^2+r^3+r^4) = 77
a(r^5+r^6+r^7+r^8+r^9) = -2464
r^5 = - 2464/77 = -32
r = -2
a(1-2+4-8+16) = 77
a = 7
4th term is 7 * (-8) = -56