Question

find an infinite geometric series whose sum is 6 and such that each term is four times the sum of all the terms that follow it

Answer 1

infinte series will be ; 24/5, 24/25, 24/125 , ......

let geometric series is ; a , ar, ar², ar³ , ar⁴ , ........∞

given, sum of infinite series is 6

i.e., a/(1 - r) = 6 [ sum of infinite series in GP is a/(1 - r)]

a = 6 - 6r

a + 6r = 6.......(1)

again, ar^(n - 1) = 4[ar^n + ar^(n + 1) + ar^(n + 2) + ...... all terms ]

⇒ar^(n - 1) = 4ar^(n - 1)[r + r² + r³ + r⁴ + ....all terms ]

⇒1/4 = r + r² + r³ + r⁴ + ...... + all terms

here definitely r < 1

so, sum of infinite terms = r/(1 - r)

⇒1/4 = r/(1 - r)

⇒1 - r = 4r

⇒r = 1/5

and putting it in equation (1),

a = 6 - 6/5 = 24/5

so, series is ; 24/5, 24/25, 24/125 ......

also read similar questions : In an infinite geometric progression,the sum of first two terms is 6 and every term is four times the sum of all the ter...

https://brainly.in/question/38168

The sum of first two terms of an infinite G.P. is 5 and each term is three times the sum of all the terms follow it . fi...

https://brainly.in/question/8080331

Answer 2

Given: An infinite geometric series whose sum is $6$ and such that each term is four times the sum of all the terms.

To Find: Find an infinite geometric series whose sum is 6 and such that each term is four times the sum of all the terms that follow it.

Step-by-step explanation:

Lets,

Take the geometric series is,

                                               $a , ar, ar^{2} , ar^{3} , ar^{4} , ............$

We know,

Sum of infinite series is given,

                                                $=\frac{a}{(1-r)}$

So,

    $\frac{a}{(1-r)}=6$

 ⇒$a=6-6r$__1

Geometric series such that each term is $4$ times the sum of all the terms,

So,

    $ar^{n-1} = 4[ar^n + ar^{n+1} + ar^{n+2}+..........]$

              $=4ar^{n-1}[r + r^{2} + r^{3}+..........]$

⇒$\frac{1}{4} =[r + r^{2} + r^{3}+..........]$

Here,  $[r + r^{2} + r^{3}+..........]$ is given,

    $=\frac{r}{1-r}$

So,

  $\frac{1}{4} =\frac{r}{1-r}$

⇒$1-r=4r$

⇒$5r=1$

∴ $r=\frac{1}{5}$

Plug the $r$ value in equation-1,

  ∴ $a=6-6\times \frac{1}{5}$

  ⇒$a=\frac{30-6}{5}$

  ∴ $a=\frac{24}{5}$

∴ $ar=\frac{24}{5}\times \frac{1}{5}=\frac{24}{25},ar^{2}=\frac{24}{5}\times \frac{1}{5^{2} } =\frac{24}{125},$

Therefore, The infinite geometric series is $\frac{24}{5},\frac{24}{25},\frac{24}{125},................$