Sequence And Series Question 503
Question: If $ a_1,,a_2,….,{a_{n+1}} $ are in A.P., then $ \frac{1}{a_1a_2}+\frac{1}{a_2a_3}+…..+\frac{1}{a_{n}{a_{n+1}}} $ is
[AMU 2002]
Options:
A) $ \frac{n-1}{a_1{a_{n+1}}} $
B) $ \frac{1}{a_1{a_{n+1}}} $
C) $ \frac{n+1}{a_1{a_{n+1}}} $
D) $ \frac{n}{a_1{a_{n+1}}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ a_1,a_2,a_3,…….,{a_{n+1}} $ are in A.P. and common difference = d Let $ S=\frac{1}{a_1a_2}+\frac{1}{a_2a_3}+……….+\frac{1}{a_{n}{a_{n+1}}} $
Þ $ S=\frac{1}{d},{ \frac{d}{a_1a_2}+\frac{d}{a_2a_3}+……+\frac{d}{a_{n}{a_{n+1}}} } $
Þ $ S=\frac{1}{d},{ \frac{a_2-a_1}{a_1a_2}+\frac{a_3-a_2}{a_2a_3}+……+\frac{{a_{n+1}}-a_{n}}{a_{n},{a_{n+1}}} } $
Þ $ S=\frac{1}{d}{ \frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}+…….+\frac{1}{a_{n}}-\frac{1}{{a_{n+1}}} } $
Þ $ S=\frac{1}{d}{ \frac{1}{a_{n}}-\frac{1}{{a_{n+1}}} }=\frac{1}{d}{ \frac{{a_{n+1}}-a_1}{a_1{a_{n+1}}} } $
Þ $ S=\frac{1}{d}( \frac{nd}{a_1{a_{n+1}}} )=\frac{n}{a_1{a_{n+1}}} $ . Trick: Check for $ n=2 $ .
BETA
