Question

The sum to infinity of a geometric series is 15 and the first term is 3. The common ratio is

Answer 1

the answer is 4/5.

S∞=a/(1-r)

15=3/(1-r)

15-15r=3

15r=12

r=4/5

Answer 2

Concept

When one term is varied by another by a common ratio, the series is referred to as a geometric progression or sequence.

The sum to an infinite geometric progression,$a,ar,ar^{2} ,...$ , is given as-

$\[S=\frac{a}{1-r}\]$

where, $r$ is the common ratio and $a$ is the initial term.

Given

With an initial term of $3$, the sum of an infinite geometric progression is $15$.

Find

We have to find the common ratio of an infinite geometric progression.

Solution

The sum to an infinite geometric progression is given as-

$\[S=\frac{a}{1-r}\]$

This is equal to 15, i.e.

$\[\frac{a}{1-r}=15\]$

Substituting the value of the initial term $\[a=3\]$, we get

$\[\frac{3}{1-r}=15\]$

$\[3=15\times \left( 1-r \right)\]$

$\[3=15-15r\]$

$\[15r=15-3\]$

$\[15r=12\]$

$\[r=\frac{12}{15}\]$

$\[r=\frac{4}{5}\]$

$\[r=0.8\]$

Hence, 0.8 is the required common ratio.

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