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If $x, 2 x+2,3 x+3$, are in G.P., then the fourth term is
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Verified Answer
The correct answer is:
$-13.5$
Given that $x, 2 x+2,3 x+3$ are in G.P.
Therefore, $(2 x+2)^2=x(3 x+3) \Rightarrow x^2+5 x+4=0$
$\Rightarrow(x+4)(x+1)=0 \Rightarrow x=-1,-4$
Now first term $a=x$
Second term $a r=2(x+1)$ $\Rightarrow r=\frac{2(x+1)}{x}$
then $4^{\text {th }}$ term $=a r^3$ $=x\left[\frac{2(x+1)}{x}\right]^3=\frac{8}{x^2}(x+1)^3$
Putting $x=-4$
We get, $T_4=\frac{8}{16}(-3)^3=-\frac{27}{2}=-13.5$
Therefore, $(2 x+2)^2=x(3 x+3) \Rightarrow x^2+5 x+4=0$
$\Rightarrow(x+4)(x+1)=0 \Rightarrow x=-1,-4$
Now first term $a=x$
Second term $a r=2(x+1)$ $\Rightarrow r=\frac{2(x+1)}{x}$
then $4^{\text {th }}$ term $=a r^3$ $=x\left[\frac{2(x+1)}{x}\right]^3=\frac{8}{x^2}(x+1)^3$
Putting $x=-4$
We get, $T_4=\frac{8}{16}(-3)^3=-\frac{27}{2}=-13.5$
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