# Question 5 and 6 Exercise 4.2

Solutions of Question 5 and 6 of Exercise 4.2 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 5

Show that the sequence $$\log a, \log (a b), \log \left(a b^2\right), \log \left(a b^3\right), \ldots$$ is an A.P. Also find its nth term.

### Solution

We first find $n$th term. Each term of the sequence is $\log$ of some number. Each log contains $a$ but the power of $b$ in first term is zero, in second term the power of $b$ is 1 and so on, therefore $$a_n=\log (a b^{n-1}).$$ We show that the given sequence is A.P. Since \begin{align}a_n&=\log(a b^{n-1}). \end{align} Now we take \begin{align} d&=a_{n+1}-a_n \\ &=\log (a b^n)-\log (a b^{n-1}) \\ &=\log \left(\dfrac{a b^n}{a b^{n-1}}\right)\\ &=\log b. \end{align} We see that the difference of consecutive terms $d$ is constant, i.e. independent of $n$.

Thus, the given sequence is in A.P.

## Question 6

Find the value of $k$, if $2 k+7,6 k-2$, $8 k-4$ are in A.P. Also find the sequence.

### Solution

Since the given terms are in A.P, \begin{align}& (6 k-2)-(2 k+7)=(8 k-4)-(6 k-2)\\ \implies & 6 k-2 k-2-7=8 k-6 k-4+2 \\ \implies & 4 k-9=2 k-2 \\ \implies & 4 k-2 k=-2+9 \\ \implies & 2 k=7\\ \implies & k=\dfrac{7}{2}.\end{align} Now the terms are: \begin{align}a_1&=2 k+7=2 \cdot \dfrac{7}{2}+7=14 \\ a_2&=6 k-2=6 \cdot \dfrac{7}{2}-2=19 \\ a_3&=8 k-4=8 \cdot \dfrac{7}{2}-4=24 .\end{align} Hence $k=\dfrac{7}{2}$ and the sequence is $14,19,24, \ldots$.

### Go To