Question
Question: The sum of \(n\)terms of the sequence \(\log a,\log ar,\log a{r^2},...\) is: A.\(\dfrac{n}{2}\log ...
The sum of nterms of the sequence loga,logar,logar2,... is:
A.2nloga2rn−1
B.nloga2rn−1
C.23nloga2rn−1
D.None of these
Solution
Hint: We first simplify the given sequence and then find its first term and common difference. To find the sum of nterms of the given sequence, use the formula , Sn=2n(2a+(n−1)d). To get the final answer, use the properties of log to simplify the answer.
Complete step by step answer:
The first term of the sequence is loga
We will first simplify the given sequence.
We know that, log(xy)=logx+logy
Therefore, we write the given sequence as, loga,loga+logr,loga+logr2,...
Also, log(xm)=mlogx
Hence, we have, loga,loga+logr,loga+2logr,...
Now, we can observe that the given sequence is an AP as logr is added to each term.
Now, we will find the common difference by subtracting the second tern from the first one.
Hence, we get common difference d as,
d=loga+logr−loga d=logr
We have to find the sum of nterms of the given sequence.
We know that the sum of n terms of a sequence is 2n(2a+(n−1)d), where a is the first term and d is the common difference.
Substitute the values of the first term and the common difference to find the sum of nterms of the sequence.
Hence, sum of n terms is
2n(2(loga)+(n−1)logr)
We can simplify the above expression using the properties of log, log(xm)=mlogx
⇒ 2n(loga2+logrn−1)
Hence, option A is correct.
Note: We have to use properties of log to simplify the given sequence such as, log(xy)=logx+logy and log(xm)=mlogx. Many students make mistakes by considering the given sequence as GP. It is after applying the properties of log, we will observe that the given sequence is an AP.