The sum of a G.P. and the sum to infinity · Sequences & series (AP, GP)

To add the first $n$ terms of a G.P., write the sum $S_n$ , then multiply every term by $r$ and write $rS_n$ underneath, shifted one place. Every term from $ar$ to $ar^{n - 1}$ is in both rows, so taking one row from the other leaves just the ends:

rS_n - S_n = ar^n - a

Take out the common factors, $S_n(r - 1) = a(r^n - 1)$ , and divide:

S_n = \frac{a(r^n - 1)}{r - 1}

Sum of a G.P.rS − S = arⁿ − a, so Sₙ = a(rⁿ − 1) ÷ (r − 1)

The same formula can be written $S_n = \dfrac{a(1 - r^n)}{1 - r}$ : top and bottom have both changed sign. Use that form when $r$ is less than 1, so both brackets are positive.

Worked example · JAMB 2011

JAMB 2011 · UTME · Q22

The second term of a geometric series is 4 while the fourth term is 16. Find the sum of the first five terms.

Write the two terms
$ar = 4$ and $ar^3 = 16$ .
Think first. Write the 2nd and 4th terms with a and r.
Divide
$r^2 = 4$ , so $r = 2$ or $r = -2$ .
Think first. What power of r is left? How many values can r have?
Try r = 2
$a = \frac42 = 2$ . $S_5 = \dfrac{2(2^5 - 1)}{2 - 1} = 2 \times 31 = 62$ .
Think first. Find a, then S₅.
Try r = −2
$a = -2$ . $S_5 = \dfrac{-2\big((-2)^5 - 1\big)}{-2 - 1} = \dfrac{-2 \times (-33)}{-3} = -22$ , not an option. The answer is 62 (B).
Think first. Is this sum one of the options?

The sum to infinity

When $r$ is between $-1$ and $1$ , each term is smaller than the one before, and $r^n$ gets closer and closer to 0 as $n$ grows. So $S_n = \dfrac{a(1 - r^n)}{1 - r}$ gets closer and closer to one number, the sum to infinity:

S_\infty = \frac{a}{1 - r} \qquad (-1 < r < 1)

Sum to infinityWith r = ½ each piece is half the gap left: S∞ = a ÷ (1 − r)

Try it

Adding a G.P. for everPick r, then add more terms

7.5S4, the first 4 terms4 ÷ 0.5 = 8S∞ = a ÷ (1 − r)0.5still to go

a = 4n = 4

Each term is a fraction of the one before, so each one adds less. The totals climb towards the dashed line but never pass it: after 4 terms there is 0.5 still to go. The line is the sum to infinity, S∞ = 4 ÷ (1 − ½) = 8.

Add terms one at a time. For $r = \frac12$ , $\frac13$ and $\frac34$ the totals creep up to the dashed line; $\frac34$ gets there slowest. For $r = -\frac12$ they jump either side of it. For $r = 2$ they run away: no sum to infinity.

Working backwards

The sum to infinity gives one equation in $a$ and $r$ . If you know one of them, solve for the other.

Worked example · JAMB 1999

JAMB 1999 · UME · Q12

The first term of a geometric progression is twice its common ratio. Find the sum of the first two terms of the progression if its sum to infinity is 8.

Write a with r
$a = 2r$
Think first. 'The first term is twice its common ratio.' Write that as an equation.
Use the sum to infinity
$\dfrac{2r}{1 - r} = 8$ , so $2r = 8 - 8r$ , $10r = 8$ and $r = \frac45$ .
Think first. Put a = 2r into a ÷ (1 − r) = 8.
Find a
$a = 2 \times \frac45 = \frac85$ .
Think first. a = 2r.
The first two terms
$a + ar = \frac85 + \frac85 \times \frac45 = \frac{40}{25} + \frac{32}{25} = \frac{72}{25}$ (C).
Think first. The second term is ar.

Your turn

JAMB 2012 · UTME · Q20

Sequences & series (AP, GP)

The sum to infinity of a geometric progression is $-\frac{1}{10}$ and the first term is $-\frac18$ . Find the common ratio of the progression.

Worked solution (try it first)

Use $S_\infty = \dfrac{a}{1 - r}$ , so $1 - r = \dfrac{a}{S_\infty}$ .
Divide: $\left(-\frac18\right) \div \left(-\frac{1}{10}\right) = \frac{10}{8}$
$= \frac54$ , so $1 - r = \frac54$ .
So $r = 1 - \frac54 = -\frac14$ , option B.

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