Question

The sum of the first two terms of a G.P is 5/2 , and the sum of the first four terms is 65/18. Find the G.P if r>0​

Answer 1

Common ratio r =
`(2)/(3)`

Sequence is described by

`a_n = (3)/(2)\cdot \left((2)/(3)\right)^(n-1)\\`

Explanation:

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

where
r = common ratio
a₁ = first term

Sum of first two terms

`S_2 = (a_1(1-r^2))/(1-r)`

Sum of first four terms:

`S_4 = (a_1(1-r^4))/(1-r)`

`(S_4)/(S_2) = a_1 \cdot (1-r^4)/(1-r) / a_1 \cdot (1-r^2)/(1-r)\\`

To divide, flip the divisor and multiply

`(S_4)/(S_2) =(a_1(1-r^4))/(1-r) * (1-r)/(a_1(1-r^2))`

The a₁ and (1-r) terms cancel out from numerator and denominator leaving

`(S_4)/(S_2) =(1-r^4)/(1-r^2) \cdots [1]`

Using the identity

`a^2 - b^2 = (a- b)(a+b)`

`1- r^4 = 1^4 - r^4 = (1^2 - r^2)(1^2+r^2) = (1-r^2)(1+r^2)`

Plugging this into equation 1 we get

`(S_4)/(S_2) =((1-r^(2))(1+r^(2)))/(1-r^(2))`

The
`1- r^2` terms cancel out leaving:

`(S_4)/(S_2) =1 + r^2`

We are given

`S_4 =(65)/(18)\\\\S_2 = (5)/(2)`

`(S_4)/(S_2) = (65)/(18) / (5)/(2)`

To divide, flip the denominator
`(5)/(2)` and multiply

`(S_4)/(S_2) = (65)/(18) * (2)/(5)\\\\= (13)/(9)`

Therefore

`1 + r^2 = (13)/(9)\\\\r^2 = (13)/(9) - 1\\\\= (13)/(9) - (9)/(9)\\\\= (4)/(9)`

`r = \sqrt{(4)/(9)}\\\\= (√(4))/(√(9))\\\\= (2)/(3)`

So the common ratio

`r = (2)/(3)`

To find the first term we have sum of first two terms = 5/2

`S_2 = (a_1(1-r^2))/((1-r)) = a_1 (1+ r)`

Plugging in knowns

`(5)/(2) = a_1(1+(2)/(3))\\\\= a_1 \cdot (5)/(3)`

Multiply both sides by 3/5 to get

`a_1 = (5)/(2) * (3)/(5)\\\\a_1 = (3)/(2)`

The nth term of a GP is

`a_n = a_1 \cdot r^(n-1)\\`

Plugging in the values obtained

`a_n = (3)/(2)\cdot \left((2)/(3)\right)^(n-1)\\`