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Hint: From the given conditions in the problem first we will try to find the first term and common ratio of the series using nth term formula which is given by ${T_n} = a{r^{n - 1}}$

Where ${T_n}$ is the nth term, a is the first term and r is the common ratio.

Given that:

The sum of first two term is $5$

$ \Rightarrow {T_1} + {T_2} = 5$

$ \Rightarrow a + ar = 5$ ………………………….. (1)

And each term is three times the succeeding term

${T_n} = 3({T_{n + 1}} + {T_{n + 2}} + ........\infty )$

Substituting the value of ${T_n}$ in term of a and r

$ \Rightarrow a{r^{n - 1}} = 3(a{r^n} + a{r^{n + 1}} + .....\infty )$

Simplifying the above equation we will get

$

\Rightarrow 1 = 3r(\dfrac{1}{{1 - r}}) \\

\Rightarrow 1 - r = 3r \\

\Rightarrow 4r = 1 \\

\Rightarrow r = \dfrac{1}{4} \\

$

Substitute the value of r in equation 1, we get

$

\Rightarrow a + a(\dfrac{1}{4}) = 5 \\

\Rightarrow \dfrac{5}{4}a = 5 \\

\Rightarrow a = 4 \\

$

First term of the G.P. $ = a = 4$

Second term of the G.P. $ = ar = 4 \times \dfrac{1}{4} = 1$

Third term of the G.P. $ = a{r^2} = 4 \times {\left( {\dfrac{1}{4}} \right)^2} = \dfrac{1}{4}$

And so on the G.P continues…..

Hence, the infinite G.P series is

$4,1,\dfrac{1}{4},\dfrac{1}{{16}},...........\infty $

Note: A G.P is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called common ratio. For solving this type of problem remember the formula of the nth term of a geometric progression series and read the conditions of the question carefully and then start solving for the unknown values whether common ratio or first term.

Where ${T_n}$ is the nth term, a is the first term and r is the common ratio.

Given that:

The sum of first two term is $5$

$ \Rightarrow {T_1} + {T_2} = 5$

$ \Rightarrow a + ar = 5$ ………………………….. (1)

And each term is three times the succeeding term

${T_n} = 3({T_{n + 1}} + {T_{n + 2}} + ........\infty )$

Substituting the value of ${T_n}$ in term of a and r

$ \Rightarrow a{r^{n - 1}} = 3(a{r^n} + a{r^{n + 1}} + .....\infty )$

Simplifying the above equation we will get

$

\Rightarrow 1 = 3r(\dfrac{1}{{1 - r}}) \\

\Rightarrow 1 - r = 3r \\

\Rightarrow 4r = 1 \\

\Rightarrow r = \dfrac{1}{4} \\

$

Substitute the value of r in equation 1, we get

$

\Rightarrow a + a(\dfrac{1}{4}) = 5 \\

\Rightarrow \dfrac{5}{4}a = 5 \\

\Rightarrow a = 4 \\

$

First term of the G.P. $ = a = 4$

Second term of the G.P. $ = ar = 4 \times \dfrac{1}{4} = 1$

Third term of the G.P. $ = a{r^2} = 4 \times {\left( {\dfrac{1}{4}} \right)^2} = \dfrac{1}{4}$

And so on the G.P continues…..

Hence, the infinite G.P series is

$4,1,\dfrac{1}{4},\dfrac{1}{{16}},...........\infty $

Note: A G.P is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called common ratio. For solving this type of problem remember the formula of the nth term of a geometric progression series and read the conditions of the question carefully and then start solving for the unknown values whether common ratio or first term.

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