# Question 3 and 4, Exercise 4.5

Solutions of Question 3 and 4 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

## Question 3

Find the sum of geometric series with $a_{1}=5$, $r=3$, $n=12$.

** Solution. **

Given: $a_{1}=5$, $r=3$, $n=12$

The formula to find the sum of $n$ terms of geometric series \[ S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r\neq 1. \] Thus \begin{align*} S_{12} &= \frac{5\left(1 - 3^{12}\right)}{1 - 3} \\ &= \frac{5\left(1 - 531441\right)}{-2} \\ &= \frac{5(-531440)}{-2} \\ &= \frac{-2657200}{-2} \\ &= 1328600 \end{align*}

Hence, the required sum is $1328600$.

## Question 4

Find the sum of the geometric series. $a_{1}=256, r=0.75, n=9$

** Solution. **

Given $a_1 = 256$, $r = 0.75$ and $n = 9$.

The formula to find the sum of $n$ terms of a geometric series is

$$S_n = \frac{a_1 \left(1 - r^n\right)}{1 - r}, \quad r \neq 1.$$
Thus,
\begin{align*}
S_9 &= \frac{256 \left(1 - (0.75)^9\right)}{1 - 0.75} \\
&= \frac{256 \left(1 - 0.075942 \right)}{0.25} \\
&= \frac{256 \times 0.924058}{0.25} \\
&= \frac{236.171}{0.25} \\
&= 944.684.
\end{align*}
Hence, the required sum is approximately $944.684$.

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