Question

Find the sum of the geometric progressions $1,-a,a²,-a^3,...n\text{terms}\left(ifa\ne -1\right)$

Answer 1

Step 1. Find the ratio of given GP:

Given G.P., $1,-a,a²,-a...n$terms

We know that the sum of $n$ terms of the G.P. is
$$\begin{gathered} S_n=\frac{A\left(R^n-1\right)}{R-1}\left[\because R>1\right] \\ \text{And}{S}_n=\frac{A\left(1-R^n\right)}{1-R}\left[\because R<1\right] \end{gathered}$$

Where $A=$ first term, $R=$ common ratio, and $n=$ number of terms

Here, $a=1$
$\begin{array}{rcl}R& =& \frac{-a}{1}=-a<1\end{array}$

Step 2: Find the sum of $n^{th}$ terms of GP:

Now, the sum of $n^{th}$ terms is given as

$\begin{array}{rcl}& \Rightarrow & S_n=\frac{1[1-\left(-a^n\right)]}{1-(-a)}\\ & =& \frac{1[1-\left(-a^n\right)]}{1+a}\end{array}$

Hence, the sum of $n^{th}$term is $\frac{\left[1-{\left(-a\right)}^n\right]}{1+a}$