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SEQUENCES AND SERIES — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSSEQUENCES AND SERIES3 Marks
Question
Find four numbers forming a geometric progression in which the third term is greater than the first term by $9$ and the second term is greater than by $4^{th}$ by $18$.
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Answer

Let the four numbers in G.P. be a, $a r, a r^2, a r^3$
$\therefore a r^2=a+9 \text { and } a r=a r^3+18$
Now, $a r^2-a=9$
$\Rightarrow a\left(r^2-1\right)=9 .$...(i)
And $a r-a r^3=18$
$\Rightarrow \operatorname{ar}\left(1-r^2\right)=18+$
$\Rightarrow-\operatorname{ar}\left(r^2-1\right)=18$...(ii)
Dividing eq. (ii) by eq. (i), we have
$\frac{-a r\left(r^2-1\right)}{a\left(r^2-1\right)}=\frac{18}{9}$
$\Rightarrow \mathrm{r}=-2$
Putting value of $r$ in eq. (i), we get
$a(4-1)=9$
$\Rightarrow a=3$
$\therefore a r=3 \times(-2)=-6$
$a r^2=3 \times(-2)^2=12 a r^2\{3\}$
$=3 \times(-2)^3=-24$
Therefore, the required numbers are $3,-6,12,-24$

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