Sequence And Series Question 586
Question: If $ x,,2x+2,,3x+3, $ are in G.P., then the fourth term is
[MNR 1981]
Options:
A) 27
B) $ -27 $
C) 13.5
D) $ -13.5 $
Show Answer
Answer:
Correct Answer: D
Solution:
Given that $ x,\ 2x+2,\ 3x+3 $ are in G.P. Therefore, $ {{(2x+2)}^{2}}=x(3x+3)\Rightarrow x^{2}+5x+4=0 $
$ \Rightarrow (x+4)(x+1)=0\Rightarrow x=-1,\ -4 $ Now first term $ a=x $ Second term $ ar=2(x+1) $
$ \Rightarrow r=\frac{2(x+1)}{x} $ then $ 4^{th} $ term $ =ar^{3} $ $ =x{{[ \frac{2(x+1)}{x} ]}^{3}}=\frac{8}{x^{2}}{{(x+1)}^{3}} $ Putting $ x=-4 $ We get $ T_4=\frac{8}{16}{{(-3)}^{3}}=-\frac{27}{2}=-13.5 $ .
BETA
