Question

The sum of first two terms of an infinite geometric series is 15 and each term is equal to the sum of all the terms following it. Find the series.

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Answer 1

Step-by-step explanation:

the first term of gp be a

common ratio r

a+ar=15

a(1+r)=15

a=15/(1+r)

t2=s(infinity)-s2

t2=a/1-r -15

r=1/2

a=10

series:-10,5,5/2,......

Answer 2

Concept:

A series of numbers in which the ratio of any term to its preceding term are equal, that series is called Geometric series (GP).

Sum of GP with infinite number of terms is, $S=\frac{a}{1-r}$ where $a$ is the first term of the series and $r$ is the common ratio of the GP.

The $n$ th term of GP is $=ar^{n-1}$

Given:

The sum of first two terms of GP with infinite terms is 15 and each term is equal to the terms following it.

Find:

The Infinite Geometric Series.

Solution:

The sum of first two terms, $S_2=15$

Let the first term and common ratio of the inifinite GP be $=a,r$ respectively.

Second term $(t_2)=ar$

Then,

$a+ar=15\Rightarrow a(1+r)=15\Rightarrow a=\frac{15}{1+r}\cdots(1)$

The total sum of the infinite GP, $S_\infty=\frac{a}{1-r}$

According to the condition,

$t_2=S_\infty-S_2$

$\Rightarrow ar=\frac{a}{1-r}-15$

$\Rightarrow \frac{15r}{1+r}=\frac{15}{(1+r)(1-r)}-15$

$\Rightarrow \frac{15r}{1+r}=\frac{15}{1-r^2}-15\\\Rightarrow\frac{r}{1+r}=\frac{1-1+r^2}{1-r^2}\\\Rightarrow\frac{r}{1+r}=\frac{r^2}{1-r^2}\\\Rightarrow\frac{1}{1+r}=\frac{r}{1-r^2}\\\Rightarrow1-r^2=r+r^2\\\Rightarrow2r^2+r-1=0\\\Rightarrow2r^2+2r-r-1=0\\\Rightarrow2r(r+1)-1(r+1)=0\\\Rightarrow(2r-1)(r+1)=0$

We know that if the product of numbers is zero, then at least one of them is zero.

Either, $2r-1=0\Rightarrow 2r=1\Rightarrow r=\frac{1}{2}$

Or, $r+1=0\Rightarrow r=-1$, it is not being taken as for it $a=\frac{15}{1+r}$ does not exist.

Putting the values of r in equation (1) we get,

when $r=\frac{1}{2}\Rightarrow a=\frac{15}{1+\frac{1}{2}}=\frac{15}{3/2}=\frac{30}{3}=10$

So the series is given by,

$=a,ar,ar^2,.....\\=10,10\times\frac{1}{2},10\times(\frac{1}{2})^2,.....\\=10,5,\frac{5}{2},......$

Hence the required infinite geometric series is $\mathbf{10,5,\frac{5}{2},.....}$

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