Question

What is the sum of a 6-term geometric series if the first term is 23 and the last term is 1,358,127? 1,395,030 1,461,460 1,527,890?

Answer 1

Last term is 1,527,890.

Explanation:

In this question a geometric series has been given with

first term A = 23

6th term = 1,358,127

Number of terms n = 6

we have to calculate the sum of this geometric series

we know
`T_(n)=a(r)^(n-1)`

`T_(6)=23.(r)^(6-1)=1358,127`

`r^(5)=(1358127)/(23)=59049`

`r = 59049^{(1)/(5) }`

r = 9

Now we know the formula to sum a geometric series is

`S=a.(r^(n-1) )/(r-1)`

S = 23 ×
`((9^(6-1)) )/((9-1))`

S = 23 ×
`((531441-1))/(8)`

S =
`(23.(531440))/(8)` = 23 × 66430

S = 1,527,890

Answer 2

The common ratio is (1358127/23)^(1/5) = 9.
The sum of 6 terms is
.. 23*(9^6 -1)/(9 -1) = 1,527,890 . . . . . . . . . . selection 3