Question

Find S9, the sum of the geometric series to the 9th term, in which a3=3.645 and a8=15

Answer 1

S₉ = 9.84

Explanation:

In a geometric series, the general formula for nth term is:

an = a₁rⁿ⁻¹

where,

an = nth term

a₁ = first term

r = common ratio

From this formula, we have:

a₃ = a₁r² = 3.645 --------- equation (1)

a₈ = a₁r⁷ = 15 ----------- equation (2)

dividing both of these equations, we get:

a₁r⁷/a₁r² = 15/3.645

r⁵ = 4.115

r = (4.115)^1/5

r = 1.33

Now, put this value in equation (1), we get:

a₁(1.33)² = 3.645

a₁ = 0.27

Now, the formula for the sum of geometric series upto nth term is:

Sn = a₁(rⁿ - 1)/(r - 1)

therefore,

S₉ = (0.27)(1.33⁹ - 1)/(1.33 - 1)

S₉ = 9.84

Answer 2

The sum of the first nine terms of the sequence is 74.44.

Explanation:

Geometric sequence concepts:

The nth term of a geometric sequence is given by the following equation.

`a_(n+1) = ra_(n)`

In which r is the common ratio.

This can be expanded for the nth term in the following way:

`a_(n) = a_(1)r^(n-1)`

In which
`a_(1)` is the first term.

Or even:

`a_(n) = a_(m)r^(n-m)`

The sum of the first n terms of a geometric sequence is given by:

`S_(n) = (a_(1)(1 - r^(n)))/(1 - r)`

Finding the common ratio:

`a_(3) = 3.645, a_(8) = 15`

`a_(n) = a_(m)r^(n-m)`

`a_(8) = a_(3)r^(8-3)`

`a_(3)r^(5) = a_(8)`

`3.645r^(5) = 15`

`r^(5) = (15)/(3.645)`

`r = \sqrt[5]{(15)/(3.645)}`

`r = 1.327`

Finding the first term:

`a_(3) = a_(1)r^(2)`

`a_(1) = (a_(3))/(r^(2))`

`a_(1) = (3.645)/((1.327)^(2))`

`a_(1) = 2.07`

Sum of the first nine terms:

`S_(9) = (2.07*(1 - (1.327)^(9)))/(1 - 1.327) = 74.44`

The sum of the first nine terms of the sequence is 74.44.