Sequence And Series Question 641
Question: If $ a_1,\ a_2,\ a_3,……………,\ a_{n} $ are in H.P., then $ a_1a_2+a_2a_3+ $ $ ……….+{a_{n-1}}a_{n} $ will be equal to
[IIT 1975]
Options:
A) $ a_1a_{n} $
B) $ na_1a_{n} $
C) $ (n-1)a_1a_{n} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Since $ a_1,\ a_2,\ a_3,\ ………a_{n} $ are in H.P. Therefore $ \frac{1}{a_1},\ \frac{1}{a_2},\ \frac{1}{a_3}…….\frac{1}{a_{n}} $ will be in A.P. Which gives $ \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 $
$ \Rightarrow $ $ \frac{a_1-a_2}{a_1a_2}=\frac{a_3-a_2}{a_2a_3}=…….=\frac{{a_{n-1}}-a_{n}}{{a_{n-1}}a_{n}}=d $
$ \Rightarrow $ $ a_1-a_2=da_1a_2 $ $ a_2-a_3=da_2a_3 $ ……………………….. ……………………….. and $ {a_{n-1}}-a_{n}=da_{n}{a_{n-1}} $ Adding these, we get $ d(a_1a_2+a_2a_3+……+a_{n}{a_{n-1}}) $ $ =(a_1+a_2+……+{a_{n-1}})-(a_2+a_3+…..+a_{n}) $ $ =a_1-a_{n} $ ?..(i) Also $ n^{th} $ term of this A.P. is given by $ \frac{1}{a_{n}}=\frac{1}{a_1}+(n-1)d\Rightarrow d=\frac{a_1-a_{n}}{a_1a_{n}(n-1)} $ Substituting this value of $ d $ in (i) $ (a_1-a_{n})=\frac{a_1-a_{n}}{a_1a_{n}(n-1)}(a_1a_2+a_2a_3+…….+a_{n}{a_{n-1}}) $ $ (a_1a_2+a_2a_3+………+a_{n}{a_{n-1}})=a_1a_{n}(n-1) $ .
BETA
