Find four numbers forming a geometric progression in which third term is greater than first term by

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8. Sequences and Series

medium

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 $4^{\text {th }}$ by $18 .$

Option A

Option B

Option C

Option D

Solution

Let a be the first term and r be the common ratio of the $G.P.$

$a_{1}=a, a_{2}=a r, a_{3}=a r^{2}, a_{4}=a r^{3}$

By the given condition,

$a_{3}=a_{1}+9 \Rightarrow a r^{2}=a+9$ ……….$(1)$

$a_{4}=a_{4}+18 \Rightarrow a r=a r^{3}+18$ ……….$(2)$

From $(1)$ and $(2),$ we obtain

$a\left(r^{2}-1\right)=9 $ ……….$(3)$

$a r\left(1-r^{2}\right)=18$ ………..$(4)$

Dividing $(4)$ by $(3),$ we obtain

$\frac{\operatorname{ar}\left(1-r^{2}\right)}{a\left(r^{2}-1\right)}=\frac{18}{9}$

$\Rightarrow-r=2$

$\Rightarrow r=-2$

Substituting the value of $r$ in $(1),$ we obtain

$4 a=a+9$

$\Rightarrow 3 a=9$

$\therefore a=3$

Thus, the first four numbers of the $G.P.$ are $3,3(-2), 3(-2)^{2},$ and $3(-2)^{3}$

i.e., $3,-6,12$ and $-24$

Standard 11

Mathematics

In a geometric progression, if the ratio of the sum of first $5$ terms to the sum of their reciprocals is $49$, and the sum of the first and the third term is $35$ . Then the first term of this geometric progression is

hard

(JEE MAIN-2014)

The sum of first three terms of a $G.P.$ is $\frac{39}{10}$ and their product is $1 .$ Find the common ratio and the terms.

medium

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easy

Let $A _{1}, A _{2}, A _{3}, \ldots \ldots$ be an increasing geometric progression of positive real numbers. If $A _{1} A _{3} A _{5} A _{7}=\frac{1}{1296}$ and $A _{2}+ A _{4}=\frac{7}{36}$, then, the value of $A _{6}+ A _{8}+ A _{10}$ is equal to

hard

(JEE MAIN-2022)

The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + ……$ will be