Sequence And Series Question 134
The natural numbers are written as follows $ 1 $ x $ \begin{matrix} 2 & 3 \\ \end{matrix} $ $ \begin{matrix} 4 & 5 & 6 \\ \end{matrix} $ $ \begin{matrix} 7 & 8 & 9 & 10 \\ . & . & . & . \\ . & . & & . \\ . & . & . & . \\ \end{matrix} $ The sum of the numbers in the $ n^{th} $ row is
Options:
A) $ \frac{n}{2}(n^{2}-1) $
B) $ \frac{n}{2}(n^{2}+1) $
C) $ \frac{2}{n}(n^{2}+1) $
D) $ \frac{2}{n}(n^{2}-1) $
Show Answer
Answer:
Correct Answer: B
Solution:
$ T_{n}=\frac{1}{2}(n^{2}-n+2), $ this is the first term of the $ n^{th} $ row. Also, the terms of the $ n^{th} $ row form an A.P. with its first term as $ \frac{1}{2}(n^{2}-n+2) $ and a common difference of 1. Hence the sum of the terms in the $ n^{th} $ row is $ p:q:r=(-1\pm \sqrt{3})K:2K:(-1\mp \sqrt{3})K $ .
BETA
