Sequence And Series Question 109
Question: If $ a_1,a_2….a_{n} $ are in H.P., then the expression $ a_1a_2+a_2a_3 $ +…+ $ a_{n}{-1}a{n} $ is equal to
Options:
A) $ n(a_1-a_{n}) $
B) $ (n-1)(a_1-a_{n}) $
C) $ na_1a_{n} $
D) $ (n-1)a_1a_{n} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ \frac{1}{a_2}-\frac{1}{a_1}=\frac{1}{a_3}-\frac{1}{a_2}=…=\frac{1}{a_{n}}-\frac{1}{{a_{n-1}}}=d(say) $ Then $ a_1a_2=\frac{a_1-a_2}{d},a_2a_3=\frac{a_2-a_3}{d},…,{a_{n-1}}a_{n}=\frac{{a_{n-1}}-a_{n}}{d} $
$ \therefore a_1a_2+a_2a_3+…+{a_{n-1}}a_{n}=\frac{a_1-a_{n}}{d} $ Also, $ \frac{1}{a_{n}}=\frac{1}{a_1}+(n-1)d $
$ \Rightarrow \frac{a_1-a_{n}}{d}=(n-1)a_1a_{n} $
BETA
