Question

If the first term of a G.P is 16 and its sum to infinity is 96/17 then find the common ratio

Answer 1

Answer:

The common ratio is $r = \frac{-11}{6}$

Given:

If the first term of a G.P is 16 and its sum to infinity is $\frac{96}{17}$

To find:

Find the common ratio

Solution:  

Let the GP series be represented as a + ar + $ar^2$ + ….

Sum = a + ar + $ar^2$ + ….

Sum $\frac{96}{17}$ , a = 16

By formula, $S\infty =\frac{a}{1-r}$

$\frac{96}{17}=\frac{16}{1-r}$

96 (1-r) = $16 \times 17$

1 – r = $\frac{272}{96}$

1 –  r = $\frac{17}{6}$

1-$\frac{17}{6}$= r

r =$\frac{-11}{6}.$

Result:

The common ratio is $r = \frac{-11}{6}$

Answer 2

The common ratio of G.P. is -11/6, if the first term of a G.P is 16 and its sum to infinity is 96/17.

Given: A G.P.

• First term a=16.
• Sum to infinity is 96/17.

To find: Find the common ratio.

Solution:

A G.P. is written as

$a,\: ar, \: a {r}^{2}, \: a {r}^{3}, ...$

Where, a is first term and r is common ratio.

Sum of infinite terms of G.P. is given by

$\bf S_{\infty} = \frac{a}{1 - r}$

if r<1

Step 1:

Put the given values in formula.

$\frac{96}{17} = \frac{16}{1 - r}$

Step 2:

Cross multiply and solve for r.

$96(1 - r) = 17 \times 16$

or

$96 - 96r = 272$

or

$- 96r = 272 - 96$

or

$- 96r = 176$

or

$r = \frac{ - 176}{96}$

or

After cancelling common factors

$\bf r = \frac{ - 11}{6}$

Common ratio (r) of G.P. is -11/6.

Learn more:

1) For a G.P.

i) if a = 2, r = 3, $\rm S_{n}$ = 242, find n.

ii) if $\rm S_{3}$ = 125, [tex]\rm S_{6}[/...

https://brainly.in/question/7783597

2) If (a-2), 8,16, are in GP then a =

https://brainly.in/question/45885832