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8. Sequences and Series
easy
If $x,\,2x + 2,\,3x + 3,$are in $G.P.$, then the fourth term is
A
$27$
B
$- 27$
C
$13.5$
D
$- 13.5$
Solution
(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 ${T_4} = \frac{8}{{16}}{( – 3)^3} = – \frac{{27}}{2} = – 13.5$.
Standard 11
Mathematics
