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PART - 2 CH - 8 Sequences and Series — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsPART - 2 CH - 8 Sequences and Series2 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 the 4th by 18.
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Answer

Let the four terms of the geometric progression be $a$, $a r, a r^2$ and $a r^3$.
According to question,
$a r^2-a=9$
$\Rightarrow \quad a\left(r^2-1\right)=9$..........(i)
and $\quad a r-a r^3=18$
$\Rightarrow \quad a r\left(1-r^2\right)=18$.........(ii)
On dividing by equation (i) in equation (ii)
$\frac{a r\left(1-r^2\right)}{a\left(r^2-1\right)}=\frac{18}{9}$
$\Rightarrow \quad-r=2 \quad \Rightarrow r=-2$
Putting value of $r$ in equation (i)
$a(4-1)=9 \quad \Rightarrow a=3$
$\begin{array}{l}\therefore \text { Required numbers are } 3,3(-2), 3(-2)^2, 3(-2)^3 \\ \quad=3-6 \cdot 12-24 \text {}\end{array}$

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