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 $14,19,24, \ldots$.
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