Sequence-And-Series Question 664
Question: If $ a_1,a_2,a_3,…..a_{n} $ are in A.P., and $ \frac{1}{a_1a_{n}}+\frac{1}{a_2{a_{n-1}}}+\frac{1}{a_3{a_{n-2}}}+…..+\frac{1}{a_{n}a_1} $ $ =K( \frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+….+\frac{1}{a_{n}} ) $ . Then K is
Options:
A) $ \frac{2}{a_1+a_{n}} $
B) $ \frac{n}{a_1+a_{n}} $
C) $ \frac{1}{a_1+a_{n}} $
D) $ \frac{n-1}{a_1+a_{n}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] We know that in an A.P. $ a_1+a_{n}=a_2+{a_{n-1}}=a_3+{a_{n-2}}= $ ???.. (i) [see the properties of A.P.]
$ \therefore \frac{1}{a_1a_{n}}+\frac{1}{a_2{a_{n-1}}}+\frac{1}{a_3{a_{n-2}}}+….+\frac{1}{a_{n}a_1} $ $ =\frac{1}{a_1+a_{n}}[ \frac{a_1+a_{n}}{a_1a_{n}}+\frac{a_1+a_{n}}{a_2{a_{n-2}}}+\frac{a_1+a_{n}}{a_3{a_{n-2}}}+….+\frac{a_1+a_{n}}{a_{n}a_1} ] $ $ =\frac{2}{a_1+a_{n}}[ \frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+……+\frac{1}{a_{n}} ] $
BETA
