Sequence And Series Question 138
Question: The sixth term of an A.P. is equal to 2, the value of the common difference of the A.P. which makes the product $ a_1a_4a_5 $ least is given by
Options:
A) $ x=\frac{8}{5} $
B) $ x=\frac{5}{4} $
C) $ x=2/3 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ a $ be the first term and $ x $ be the common difference of the A.P. Then $ a+5x=2 $
$ \Rightarrow $ $ a=2-5x $ Let $ P=a_1a_4a_5=a,(a+3x),(a+4x) $ $ =(2-5x)(2-2x)(2-x)=2(-5x^{3}+17x^{2}-16x+4) $ Now $ \frac{dP}{dx}=0 $
$ \Rightarrow $ $ x=\frac{8}{5},\ \frac{2}{3} $ . Clearly, $ \frac{d^{2}P}{dx^{2}}>0 $ for $ x=\frac{2}{3} $ Hence $ P $ is least for $ x=\frac{2}{3} $ .
BETA
