====== 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. GOOD

=====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. GOOD
====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 $14,19,24, \ldots$. GOOD

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