Question

Sum of infinite terms of a GP is 12. If the first term is 8, what is the 4th term of this GP?
\\frac{8}{27}\\)
\\frac{4}{27}\\)
\\frac{8}{20}\\)
\\frac{1}{3}\\)

Answer 1

Answer:

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Answer 2

QUESTION:

The correct question should be:

The sum of infinite terms of a GP is 12. If the first term is 8, what is the 4th term of this GP?

$A. \frac{8}{27}$

$B. \frac{4}{27}$

$C. \frac{8}{20}$

$D. \frac{1}{3}$

ANSWER:

The $4^{th}$ term of this GP is $\frac{8}{27}$. (option A)

Given,

The sum of infinite terms of a GP = 12,

The first term of the GP = 8.

To find,

4th term of this GP.

Solution,

The $n^{th}$ term a geometric progression, or GP, that is $a_n$, is given by,

$a_n=ar^{(n-1)}$     ...(2)

Where,

$a$ = first term,

r = common ratio, which is given by the ratio of a term to its preceding term, that is, $r=\frac{a_n}{a_{n-1}}$

And, the sum of infinite terms of a geometric progression ($S_{\infty}$), is given by the formula,

$S_{\infty} =\frac{a}{1-r}$     ...(2)

Here, the sum given is, $S_{\infty}=12,$ and,

the first term is, a = 8. So,

$12=\frac{8}{1-r}$

Rearranging and simplifying, we get,

$1-r=\frac{8}{12}$

$\implies r =1-\frac{8}{12} =\frac{12-8}{12}=\frac{4}{12}$

$\implies r =\frac{1}{3}$

Using the above value of r, and the equation (1), the $4^{th}$ term can be determined as follows.

$r =\frac{1}{3},$

n = 4,

a = 8.

Thus,

$a_4=ar^{4-1}$

Substituting the above values of r and a,

$\implies a_4=(8)(\frac{1}{3} )^{3}$

$\implies a_4=\frac{8}{27}$

Therefore, the $4^{th}$ term of this GP will be $\frac{8}{27}$. (option A)