# Question 5 and 6, Exercise 4.5

Solutions of Question 5 and 6 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 5

Find the sum of the geometric series. $a_{1}=7, r=2, n=14$

** Solution. **

Given $a_1 = 7$, $r = 2$ and $n = 14$.

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_{14} &= \frac{7 \left(1 - 2^{14}\right)}{1 - 2} \\
&= \frac{7 \left(1 - 16384\right)}{-1} \\
&= \frac{7 \times (-16383)}{-1} \\
&= 7 \times 16383 \\
&= 114681.
\end{align*}
Hence, the required sum is $114681$.

## Question 6

Find the sum of the geometric series. $a_{1}=12, a_{5}=972, r=-3$

** Solution. **

Given $a_1 = 12$, $a_5 = 972$ and $r = -3$.

Now, we can use \(r = -3\) (as given) to find the number of terms, \(n\):
To find $n$, we use $$a_n = a_1 \cdot r^{n-1}$$
\begin{align*}
972 &= 12 (-3)^{n-1}\\
81 &= (-3)^{n-1}\\
(-3)^{n-1} &= 3^{4}\\
\implies n-1 &= 4 \\
n &= 5\end{align*}
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_5 &= \frac{12 \left(1 - (-3)^{5}\right)}{1 - (-3)} \\
&= \frac{12 \left(1 + 243\right)}{4} \\
&= \frac{12 \times 244}{4} \\
&= 12 \times 61 \\
&= 732.
\end{align*}
Hence, the required sum is $4732$.

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