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Question
Find the sum of the following geometric series:
1, −a, a2, −a3, ....to n terms (a ≠ 1)
Solution
Common Ratio = r = `(-a)/1 = -a`
∴ Sum of GP for n terms = `(a(r^n - 1))/(r - 1)` ...(i)
⇒ a = 1, r = −a, n = n
∴ Substituting the above values in (1) we get
⇒ `(1((-a)^n - 1))/(-a-1)`
⇒ `(-1((-a)^n - 1))/(a+1)`
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