Question

find sum of Geometric Series to n terms​

Answer 1

EXPLANATION.

Sum of the series,

⇒ 1 - 1/3 + 1/3² - 1/3³ +   to n terms.

As we know that,

First term = a = 1.

Common ratio = r = -1/3.

As we know that,

⇒ S∞ = a/1 - r.

⇒ S∞ = 1/1 - (-1/3).

⇒ S∞ = 1/1 + 1/3.

⇒ S∞ = 1/4/3.

⇒ S∞ = 3/4.

                                                                                         

MORE INFORMATION.

Geometrical Mean (G.M).

If G is the G.M between two numbers a and b then,

G² = ab ⇒ G = √ab.

n GM's between two given numbers a and b.

$\sf \implies r = \bigg(\dfrac{b}{a}\bigg) ^{\dfrac{1}{n + 1}}$

Then G₁ = ar, G₂ = ar², G₃ = ar³  ⇒ Gₙ = arⁿ  Or  Gₙ = b/r  and,

$\sf \implies G_K = a\bigg(\dfrac{b}{a}\bigg)^{\dfrac{1}{n + 1} }$

$\sf \implies where \ G_k \ is \ K^{th} \ G.M \ between\ a \ and \ b.$

Product of n G.M's inserted between a and b is $(ab)^{\dfrac{n}{2} }$

Supposition of term in G.P.

(1) = Three terms as = a/r, a, ar.

(2) = Four terms as = a/r³, a/r, ar, ar³.

(3) = Five terms as = a/r², a/r, a, ar, ar².

Answer 2

$\huge \rm{ \underline {\underline{To \: find : }}}$

• The sum of the given geometric series to n terms;

$\huge \rm \red{ \underline {\underline{ \color{black}Solution}}}$

Given series:-

$\rightarrow \sf1 \frac{1}{3} + \frac{1}{ {3}^{2} } - \frac{1}{ {3}^{2} } + \: to \: n \: terms$

✮We know that;

First term = a = 1

Common ratio = r = $\sf \frac{ - 1}{3}$

✮We know that;

$\rightarrow \sf S \infty = \large \frac{a}{1} - r$

$\sf \rightarrow S \infty = \large\frac{1}{1} - \frac{( - 1)}{3}$

$\sf \rightarrow S \infty = \large \frac{1}{1} + \frac{1}{3}$

$\rightarrow \sf S \: \infty =\large \frac{1}{ \frac{4}{3} }$

$\sf \implies S \infty =\large \frac{3}{4}$