Question

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

Do not round your answer.
7+21 +63 +...

Answer 1

Sum of 9terms = 68,887

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

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

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=(21)/(7)=3`

The sum of a geometric sequence is given by:

`S_n=(a_1(r^n-1))/(r-1)` when
`r>1`

and

`S_n=(a_1(1-r^n))/(1-r)` when
`r<1`

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=(7(3^9-1))/(3-1)`

`S_9=(7(19683-1))/(3-1)`

`S_9=(7(19682))/(2)`

`S_9=(137774)/(2)`

∴
`S_9=68887`

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