Question

An exponential sequence (G.P) has a positive common ratio. If the sum to infinity of the sequence is 25 and the sum of the first 2 terms is 16, find the;

i)Fifth term ii)Sum of the first 4 terms of the sequence ?

Answer 1

Let
`a` be the first term in the sequence, and
`r` the common ratio between consecutive terms. Let
`a_n` denote the
`n`-th term in the sequence. The first several terms are

`\{a, ar, ar^2, ar^3, \ldots\}`

and so the
`n`-th term is given by
`a_n = ar^(n-1)`.

Let
`S_N` denote the
`N`-th partial sum of the series,

`\displaystyle S_N = \sum_(n=1)^N ar^(n-1) = a + ar + ar^2 + \cdots + ar^(N-1)`

Multiply both sides by
`r`.

`r S_N = ar + ar^2 + ar^3 + \cdots + ar^N`

Subtract this from, then solve for,
`S_N`.

`S_N - r S_N = a - ar^N \implies S_N = (a(1 - r^N))/(1 - r)`

We know the infinite series converges, so
`|r|<1`, which means the
`S_N \to \frac a{1-r}` as
`N\to\infty`. And since we know the infinite sum is 25, we have

`\frac a{1-r} = 25 \implies a + 25r = 25`

Meanwhile, the sum of the first 2 terms is

`a + ar = 16`

Solve for
`r`.

`a + ar = 16 \implies ar = 16 - a \\\\ \implies r = \frac{16}a - 1`

Substitute this into the other equation and solve for
`a`, then again for
`r`.

`a + 25\left(\frac{16}a - 1\right) = 25 \implies a + \frac{400}a - 25 = 25 \\\\ \implies a - 50 + \frac{400}a = 0 \\\\ \implies a^2 - 50a + 400 = 0 \\\\ \implies (a-10) (a-40) = 0 \\\\ \implies a = 10 \text{ or } a = 40`

`\implies r = (16)/(10) - 1 = \frac35 \text{ or } r = (16)/(40) - 1 = -\frac35`

We're given that the ration is positive, so
`r=\frac35` and
`a=10`.

i) The fifth term in the sequence is

`ar^(5-1) = 10 \left(\frac35\right)^4 = (162)/(125) = 1.296`

ii) The sum of the first 4 terms is

`\displaystyle S_4 = \sum_(n=1)^4 ar^(n-1) = (a(1 - r^4))/(1 - r) = (10\left(1 - \left(\frac35\right)^4\right))/(1 - \frac35) = (544)/(25) = 21.76`