# Question 4 & 5 Exercise 4.4

Solutions of Question 4 & 5 of Exercise 4.4 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 4

How many terms are there in a geometric sequence in which the first and the last terms are 16 and $\dfrac{1}{64}$ respectively and $r=\dfrac{1}{2}$ ?

### Solution

First term of the sequence $a_1=16$

Last term of the sequence $a_n=\dfrac{1}{64}$ Common ratio $r=\dfrac{1}{2}$

We have to find $n$

We know that $$a_n=a_1 r^{n-1} \quad \text{then}$$
\begin{align}\dfrac{1}{64}&=16(\dfrac{1}{2})^{n-1} \\
\Rightarrow(\dfrac{1}{2})^{n-1}&=\dfrac{1}{64 \times 16}=\dfrac{1}{1024} \\
\Rightarrow(\dfrac{1}{2})^{n-1}&=\dfrac{1}{2^{10}}=(\dfrac{1}{2})^{10} \\
\Rightarrow n-1&=10 \text { or } n=11\end{align}
Hence the total number of terms are $11$ in sequence.

## Question 5

Find $x$ so that $x+7, x-3, x-8$ forms a three terms geometric sequence in the given order. Also give the sequence.

### Solution

Since $a_1=x+7$, $a_2=x-3, a_3=x-8$ forms a three terms geometric sequence,

\begin{align}\therefore \dfrac{a_2}{a_1}&=\dfrac{a_3}{a_2} \\
\Rightarrow \dfrac{x-3}{x+7}&=\dfrac{x-8}{x-3} \\
\Rightarrow(x-3)^2&=(x-8)(x+7) \\
\Rightarrow x^2-6 x+9&=x^2-x-56 \\
\Rightarrow-6 \cdot+x&=-56-9 \\
\Rightarrow-5 x&=-65 \\
\Rightarrow x&=\dfrac{-65}{-5}=13 .\end{align}
Thus the elements of the sequence are

\begin{align}x+7&=13+7=20 \\
x-3&=13+3=10 \quad\text { and } \\
x-8&=13-8=5\\
x=13;20,10,5\end{align}

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