Find the sum of the geometric series 3 + 6 + 12 + … + 1536.

Question:

Find the sum of the geometric series 3 + 6 + 12 + … + 1536.

Solution:

Tn represents the $n^{\text {th }}$ term of a G.P. series.

$r=6 \div 3=2$

$\mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1}$

$\Rightarrow 1536=3 \times 2^{n-1}$

$\Rightarrow 1536 \div 3=2^{n} \div 2$

$\Rightarrow 1536 \div 3 \times 2=2^{n}$

$\Rightarrow 1024=2^{n}$

$\Rightarrow 2^{10}=2^{n}$

∴ n = 10

Sum of a G.P. series is represented by the formula, $S_{n}=a \frac{r^{n}-1}{r-1}$ hen r>1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 3

r = 2 

n = 10 terms

$\therefore \mathrm{S}_{\mathrm{n}}=3 \times \frac{2^{10}-1}{2-1}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=3 \times(1024-1)$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=3 \times 1023$

$\therefore \mathrm{S}_{\mathrm{n}}=3069$

 

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