Question

Find the sum of the first 9 terms in the following geometric series.

Answer 1

Answer:

Sum of 9terms = 68,887

Step-by-step explanation:

Sum nth term of a GP series is Sn = a(r^n -1)/(r-1)

where a = first term

r = common ratio = Tn/Tn-1

n = nth of term

Therefore for 7,21 ,63 +...

a = 7

r = 21/7 = 3

I.e

Sum of 9 terms = 7 x (3^9-1)/(3-1)

=7 x (19683-1)/2

7 x 19682/2

= 7 x 9841

= 68,887

Sum of 9terms = 68,887

Answer 2

Answer:

The sum of first 9 terms of the given sequence = 68887

Step-by-step explanation:

Given sequence:

7+21+63......

The given sequence is a geometric sequence as the successive numbers bear a common ratio.

The ratio can be found out by dividing a number by the number preceding it.

For the given geometric sequence common ratio $r$ can be given as:

$r=\frac{21}{7}=3$

The sum of a geometric sequence is given by:

$S_n=\frac{a_1(r^n-1)}{r-1}$       when $r>1$

and

$S_n=\frac{a_1(1-r^n)}{1-r}$      when $r$

where, $S_n$ represents sum of $n$terms, $n$ representing number of terms and $r$ represents common ratio and $a_1$ represents the first term.

Since for the given geometric sequence has a common ratio =3 which is >1, so we will use the first formula for sum to calculate the sum of first 9 terms.

Plugging in the values to find sum of first 9 terms.

$S_9=\frac{7(3^9-1)}{3-1}$  

$S_9=\frac{7(19683-1)}{3-1}$  

$S_9=\frac{7(19682)}{2}$

$S_9=\frac{137774}{2}$

∴ $S_9=68887$  

Thus sum of first 9 terms of the given sequence = 68887 (Answer)