# How do you find the sum of a geometric series for which a1 = 48, an = 3, and r = -1/2?

Jun 19, 2018

${S}_{5} = 33$

#### Explanation:

We know that ,

$\text{(1)The nth term of a geometric series is :}$

color(blue)(a_n=a_1(r)^(n-1), where, a_1=1^(st)term and r=common  $r a t i o$

$\text{(2) The sum of first n term of the geometric series is :}$

color(red)(S_n=(a_1(1-r^n))/(1-r) ,where, r!=1

Here, ${a}_{1} = 48 , {a}_{n} = 3 \mathmr{and} r = \left(- \frac{1}{2}\right)$

Using $\left(1\right)$ we get

${a}_{n} = 48 {\left(- \frac{1}{2}\right)}^{n - 1} = 3$

$\implies {\left(- \frac{1}{2}\right)}^{n - 1} = \frac{3}{48} = \frac{1}{16}$

$\implies {\left(- \frac{1}{2}\right)}^{n - 1} = {\left(- \frac{1}{2}\right)}^{4}$

$\implies n - 1 = 4$

$\implies n = 5$

Now ,using $\left(2\right)$ $\text{we get, "color(violet)"sum of first five terms is :}$

S_5=(48[1-(-1/2)^5])/(1-(-1/2)

$\implies {S}_{5} = \frac{48 \left[1 - \left(- \frac{1}{32}\right)\right]}{1 + \frac{1}{2}}$

$\implies {S}_{5} = \frac{48 \left[1 + \frac{1}{32}\right]}{\frac{3}{2}}$

$\implies {S}_{5} = \frac{96 \left(\frac{33}{32}\right)}{3}$

$\implies {S}_{5} = 33$
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Note:

The sum of first n term of this series is :

S_n=(48[1-(-1/2)^n])/(1-(-1/2)

${S}_{n} = \frac{48 \left[1 - {\left(- \frac{1}{2}\right)}^{n}\right]}{\frac{3}{2}}$

${S}_{n} = \frac{2}{3} \times 48 \left[1 - {\left(- \frac{1}{2}\right)}^{n}\right]$

${S}_{n} = 32 \left[1 - {\left(- \frac{1}{2}\right)}^{n}\right]$