Find the sum to indicated number of terms in the geometric progression given:
$1, - a, a^{2}, - a^{3}, . . . . n$ terms (if $a \neq - 1$ )

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Solution

first term $= a =!$ ; common ratio $= - a = r$
Sum = $\frac{a (r^{m} - 1)}{r - 1} = \frac{(1) ((- a)^{m} - 1)}{- a - 1} = \frac{1 - (- a)^{m}}{1 + a}$

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Find the sum to indicated number of terms in the geometric progression given:1,−a,a2,−a3,....n terms (if a≠−1)

first term =a=!; common ratio =−a=rSum = a(rm−1)r−1=(1)((−a)m−1)−a−1=1−(−a)m1+a

Find the sum to indicated number of terms in the geometric progression given:
$1, - a, a^{2}, - a^{3}, . . . . n$ terms (if $a \neq - 1$ )

first term $= a =!$ ; common ratio $= - a = r$
Sum = $\frac{a (r^{m} - 1)}{r - 1} = \frac{(1) ((- a)^{m} - 1)}{- a - 1} = \frac{1 - (- a)^{m}}{1 + a}$