Sequence And Series Question 234
Question: If 9 A.M.’s and H.M.’s are inserted between the 2 and 3 and if the harmonic mean $ H $ is corresponding to arithmetic mean $ A $ , then $ A+\frac{6}{H}= $
[ISM Dhanbad 1987]
Options:
A) 1
B) 3
C) 5
D) 6
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ A_{j},\ H_{j}, $ where $ j=1,\ 2,\ 3,…….9 $ denote the 9 A.M.?s and H.M.?s between 2 and 3. Then $ 2,A_1,A_2……..A_9,3 $ are in A.P. Let $ d $ be the common difference of this A.P. Then $ -d=\frac{a_1-a_{2n}}{2n-1} $
$ \Rightarrow $ $ d=\frac{1}{10} $ If $ A $ denotes the $ J^{th} $ arithmetic mean, then $ A=2+jd=2+( \frac{j}{10} ) $ Again $ 2,\ H_1,\ H_2…….,H_9,\ 3 $ will be in H.P. $ i.e. $ , $ \frac{1}{2},\frac{1}{H_1},\frac{1}{H_2},……\frac{1}{H_9},\ \frac{1}{3} $ will be in A.P. Let $ D $ be the common difference of this A.P. Then $ \frac{1}{3}=\frac{1}{2}+10D $
$ \Rightarrow $ $ D=-\frac{1}{60} $ If $ H $ be the $ J^{th} $ harmonic mean, then $ \frac{1}{H}=\frac{1}{2}+jD=\frac{1}{2}-\frac{j}{60} $
$ \therefore $ $ A+\frac{6}{H}=2+\frac{j}{10}+6( \frac{1}{2}-\frac{j}{60} )=5+\frac{j}{10}-\frac{j}{10}=5 $ . Aliter: As we know $ A_{m}=2+\frac{m(3-2)}{9+1}=2+\frac{m}{10} $ and $ \frac{1}{H_{m}}=\frac{1}{2}+\frac{m(2-3)}{2\times 3(9+1)}=\frac{1}{2}-\frac{m}{60} $
$ \therefore $ $ A_{m}+6\times \frac{1}{H_{m}}\ \ i.e.\ \ A+\frac{6}{H}=2+\frac{m}{10}+3-\frac{m}{10}=5 $ .
BETA
