If x,,2x + 2,,3x + 3,are in G.P., then the fo | vedclass

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(d) 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 +

Vedclass · Accepted Answer

$- 13.5$

Vedclass · Answer

(d) 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 $ = a{r^3}$$ = x{\left[ {\frac{{2(x + 1)}}{x}} \right]^3} = \frac{8}{{{x^2}}}{(x + 1)^3}$

Putting $x = - 4$

We get ${T_4} = \frac{8}{{16}}{( - 3)^3} = - \frac{{27}}{2} = - 13.5$.

If $x,\,2x + 2,\,3x + 3,$are in $G.P.$, then the fourth term is

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