Question

If sequence $\log a,\log ab,\log a{{b}^{2}},\log a{{b}^{3}}......$ in A.P, then find the ${{n}^{th}}$ term and sum of the ${{n}^{th}}$term

Answer 1

Hint: In this question we have been given with an arithmetic progression of logarithmic terms for which we have to find the ${{n}^{th}}$ term and sum of $n$ terms. We will solve this question by first writing down the arithmetic progression and then we will find $a$ which is the first term in the progression and $d$ which is the common difference in the progression. we will then use the formula ${{a}_{n}}=a+\left( n-1 \right)d$ to find ${{a}_{n}}$ which is the ${{n}^{th}}$ term and use the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ to find ${{S}_{n}}$ which gives the sum of the $n$ terms.

Complete step-by-step solution:
We have the arithmetic progression given to us as:
$\Rightarrow \log a,\log ab,\log a{{b}^{2}},\log a{{b}^{3}}......$
From the above sequence, we can see that the first term of the sequence is $\log a$ therefore, we can write:
$\Rightarrow a=\log a$
Now to find the common difference we will subtract two consecutive terms. on subtracting the first term from the second term in the sequence, we get:
$\Rightarrow \log ab-\log a$
Now we know that $\log p-\log q=\log \left( \dfrac{p}{q} \right)$ therefore, on substituting, we get:
$\Rightarrow \log \left( \dfrac{ab}{a} \right)$
On simplifying the fraction, we get:
$\Rightarrow \log b$
which is the common difference therefore, we can write:
$\Rightarrow d=\log b$
Now we know that ${{n}^{th}}$ term in an arithmetic progression is given by the formula ${{a}_{n}}=a+\left( n-1 \right)d$ therefore, on substituting the values, we get:
$\Rightarrow {{a}_{n}}=\log a+\left( n-1 \right)\log b$
We know the property that $p\log q=\log {{q}^{p}}$ therefore, we get:
$\Rightarrow {{a}_{n}}=\log a+\log {{b}^{n-1}}$
Now we know that $\log p+\log q=\log pq$ therefore, we get:
$\Rightarrow {{a}_{n}}=\log \left( a{{b}^{n-1}} \right)$, which is the required ${{n}^{th}}$ term.
Now the sum of $n$ terms in an arithmetic progression is given by the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.
On substituting the value of $a$ and $d$, we get:
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2\log a+\left( n-1 \right)\log b \right]$
We know the property that $p\log q=\log {{q}^{p}}$ therefore, we get:
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ \log {{a}^{2}}+\log {{b}^{n-1}} \right]$
Now we know that $\log p+\log q=\log pq$ therefore, we get:
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\log \left( {{a}^{2}}{{b}^{n-1}} \right)$, which is the required sum of $n$ terms.

Note: In this question we have been given with an arithmetic progression therefore, the formulas relating to an A.P should be remembered. The various properties of logarithm should be remembered while doing these types of sums. It is to be noted that there also exists G.P which is called as geometric progression.