Sum of Geometric Series

When the terms of a geometric sequence are joined by plus signs, we get a geometric series. Each term comes from multiplying the one before it by the same number, called the common ratio \( r \). The first term is written \( a \), and the \( n \)th term is \( t_n = ar^{n-1} \). All of this you already know. The new question is how to add a long run of such terms quickly.

Here is a real situation. During the Covid 19 pandemic, workers in the medicine industry were in great demand. One worker started on a salary of Rs. 10,000 a month, and the company decided to raise it by 10% every month. How much does he earn altogether in the first 5 months?

Adding 10% of a number to itself is the same as multiplying the number by 1.1. So each month's salary is 1.1 times the salary of the month before.

The five salaries form a geometric series with first term \( a = 10000 \) and common ratio \( r = \frac{11000}{10000} = 1.1 \). Adding them one by one gives the total:

\[ \begin{aligned} S_5 &= 10000 + 11000 + 12100 \\ &\quad + 13310 + 14641 \\ &= 61051 \end{aligned} \]

So he earns Rs. 61,051 in 5 months. Plain adding worked here because there were only five terms. With 20 or 50 terms it would take far too long and invite mistakes, so we need a shortcut.

The shortcut rests on one clever move. Multiply the whole series by the common ratio, then subtract the original series from it. Watch it work on the salaries.

Step 1: Writing the series, and calling it equation (i):

\[ \begin{aligned} S_5 &= 10000 + 11000 + 12100 \\ &\quad + 13310 + 14641 \end{aligned} \]

Step 2: Multiplying both sides by \( r = 1.1 \). Each salary turns into the next month's salary, and a new term, 16105.1, appears at the end. Call this equation (ii):

\[ \begin{aligned} 1.1\,S_5 &= 11000 + 12100 + 13310 \\ &\quad + 14641 + 16105.1 \end{aligned} \]

Step 3: Subtracting equation (i) from equation (ii). The four middle numbers are in both lines, so they cancel, leaving only the last term of (ii) and the first term of (i):

\[ 1.1\,S_5 - S_5 = 16105.1 - 10000 \]

Step 4: Simplifying both sides:

\[ 0.1\,S_5 = 6105.1 \]

Step 5: Dividing both sides by 0.1:

\[ S_5 = 61051 \]

The same answer, with no long addition. Look at step 3 again. The number 16105.1 is \( 10000 \times (1.1)^5 \), so the working really says:

\[ S_5 = \frac{10000\{(1.1)^5 - 1\}}{1.1 - 1} \]

Its shape is: first term times (the ratio raised to the number of terms, minus 1), all over (the ratio minus 1). That same shape works for every geometric series.

Now do exactly the same thing with letters. Take a geometric series with first term \( a \), common ratio \( r \) and \( n \) terms.

Step 1: Writing the sum of the first n terms, equation (i):

\[ S_n = a + ar + ar^2 + \dots + ar^{n-1} \]

Step 2: Multiplying every term by \( r \). Each power of \( r \) goes up by one, so the last term becomes \( ar^n \). Call this equation (ii):

\[ \begin{aligned} rS_n &= ar + ar^2 + \dots \\ &\quad + ar^{n-1} + ar^n \end{aligned} \]

Step 3: Subtracting (i) from (ii). Every term from \( ar \) up to \( ar^{n-1} \) is in both lines and cancels:

\[ rS_n - S_n = ar^n - a \]

Step 4: Taking out the common factors, \( S_n \) on the left and \( a \) on the right:

\[ (r - 1)S_n = a(r^n - 1) \]

Step 5: Dividing both sides by \( r - 1 \):

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

The formula just found is the one to use when \( r > 1 \), because then \( r^n - 1 \) and \( r - 1 \) are both positive and the working stays simple.

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

When \( r < 1 \), multiply the top and the bottom by \( -1 \). This does not change the value of the fraction; it only turns both brackets round. For a ratio between 0 and 1 it keeps both brackets positive:

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

Sometimes the last term is given instead of the number of terms. Since \( t_n = ar^{n-1} \), we have \( ar^n = ar^{n-1} \times r = t_n r \). Putting this into the first formula:

\[ S_n = \frac{ar^n - a}{r - 1} = \frac{t_n r - a}{r - 1} \]

Example: Find the sum of the first 5 terms of the geometric series \( 1 + 3 + 9 + 27 + \dots \)

Reading the series: the first term is \( a = 1 \), the common ratio is \( r = \frac{3}{1} = \frac{9}{3} = 3 \), and we want \( n = 5 \) terms.

Step 1: Choosing the formula. Since \( r = 3 > 1 \), we use:

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

Step 2: Putting in \( a = 1 \), \( r = 3 \) and \( n = 5 \):

\[ S_5 = \frac{1(3^5 - 1)}{3 - 1} \]

Step 3: Working out \( 3^5 = 243 \):

\[ S_5 = \frac{243 - 1}{2} = \frac{242}{2} \]

Step 4: Dividing:

\[ S_5 = 121 \]

Check: \( 1 + 3 + 9 + 27 + 81 = 121 \). The formula and plain adding agree, so the sum of the first 5 terms is 121.

Example: Find the sum of the first 5 terms of the geometric series \( 2 + 1 + \frac{1}{2} + \frac{1}{4} + \dots \)

Here \( a = 2 \), \( r = \frac{1}{2} \) and \( n = 5 \). The terms are getting smaller, which tells us \( r < 1 \).

Step 1: Choosing the formula for \( r < 1 \):

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

Step 2: Putting in the values:

\[ S_5 = \frac{2\{1 - (\frac{1}{2})^5\}}{1 - \frac{1}{2}} \]

Step 3: Working out \( (\frac{1}{2})^5 = \frac{1}{32} \) and \( 1 - \frac{1}{2} = \frac{1}{2} \):

\[ S_5 = \frac{2(1 - \frac{1}{32})}{\frac{1}{2}} \]

Step 4: Dividing by \( \frac{1}{2} \) is the same as multiplying by 2:

\[ S_5 = 4\left(1 - \frac{1}{32}\right) = 4 \times \frac{31}{32} \]

Step 5: Simplifying:

\[ S_5 = \frac{31}{8} \]

As a mixed number this is \( 3\frac{7}{8} \). Check: \( 2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = 3\frac{7}{8} \).

A negative common ratio is also less than 1, so the same formula applies. In such a series the signs of the terms take turns. Example: Find the sum of the first 5 terms of \( 2 - 6 + 18 - 54 + \dots \) Here \( a = 2 \) and \( r = \frac{-6}{2} = -3 \).

Step 1: Putting \( a = 2 \), \( r = -3 \) and \( n = 5 \) into the formula for \( r < 1 \):

\[ S_5 = \frac{2\{1 - (-3)^5\}}{1 - (-3)} \]

Step 2: Working out \( (-3)^5 = -243 \), since an odd power of a negative number is negative:

\[ S_5 = \frac{2(1 + 243)}{4} = \frac{2 \times 244}{4} \]

Step 3: Simplifying:

\[ S_5 = 122 \]

Check: \( 2 - 6 + 18 - 54 + 162 = 122 \).

Example: Find the sum of the geometric series \( 3 + 6 + 12 + 24 + \dots + 768 \)

This series stops at 768, but we are not told how many terms it has. We know \( a = 3 \), \( r = \frac{6}{3} = 2 \) and the last term \( t_n = 768 \). Because the last term is given, the third form of the formula saves us from counting terms at all.

Step 1: Writing the formula that uses the last term:

\[ S_n = \frac{t_n r - a}{r - 1} \]

Step 2: Putting in \( t_n = 768 \), \( r = 2 \) and \( a = 3 \):

\[ S_n = \frac{768 \times 2 - 3}{2 - 1} \]

Step 3: Multiplying, then subtracting:

\[ S_n = \frac{1536 - 3}{1} = 1533 \]

So the sum is 1533, and we never needed to find out that the series has 9 terms.

The formula links four quantities: \( S_n \), \( t_n \), \( r \) and \( a \). Whenever three of them are known, the fourth can be found.

Example: A geometric series has first term 7, last term 448, and the sum of all its terms is 889. Find its common ratio. Here \( a = 7 \), \( t_n = 448 \), \( S_n = 889 \), and \( r \) is unknown.

Step 1: Putting the known values into \( S_n = \frac{t_n r - a}{r - 1} \):

\[ 889 = \frac{448r - 7}{r - 1} \]

Step 2: Multiplying both sides by \( r - 1 \) to clear the fraction:

\[ 889r - 889 = 448r - 7 \]

Step 3: Collecting the \( r \) terms on the left and the numbers on the right:

\[ 889r - 448r = 889 - 7 \]

Step 4: Simplifying both sides:

\[ 441r = 882 \]

Step 5: Dividing both sides by 441:

\[ r = \frac{882}{441} = 2 \]

The common ratio is 2. Check: \( 7 + 14 + 28 + 56 + 112 + 224 + 448 = 889 \).

Example: The third and sixth terms of a geometric series are 27 and 729. Find the sum of its first 10 terms.

The sum formula needs \( a \) and \( r \), and neither is given. Each known term gives one equation through \( t_n = ar^{n-1} \), and two equations are enough to find both.

Step 1: Writing the third term, equation (i):

\[ t_3 = ar^2 = 27 \]

Step 2: Writing the sixth term, equation (ii):

\[ t_6 = ar^5 = 729 \]

Step 3: Dividing (ii) by (i). The \( a \) cancels and the powers of \( r \) subtract:

\[ \frac{ar^5}{ar^2} = \frac{729}{27} \]

Step 4: Simplifying both sides:

\[ r^3 = 27 = 3^3 \]

Step 5: The powers match, so comparing the bases:

\[ r = 3 \]

Step 6: Putting \( r = 3 \) into (i) to find \( a \):

\[ a \times 3^2 = 27 \]

Step 7: Dividing both sides by 9:

\[ a = 3 \]

Step 8: Since \( r > 1 \), using \( S_n = \frac{a(r^n - 1)}{r - 1} \) with \( n = 10 \):

\[ S_{10} = \frac{3(3^{10} - 1)}{3 - 1} \]

Step 9: Working out \( 3^{10} = 59049 \):

\[ S_{10} = \frac{3 \times 59048}{2} \]

Step 10: Simplifying:

\[ S_{10} = 3 \times 29524 = 88572 \]

So the sum of the first 10 terms is 88,572.

Example: How many terms of the geometric series \( 64 + 32 + 16 + \dots \) will have a sum of \( \frac{255}{2} \)?

This time the unknown is \( n \), and it sits in a power. Here \( a = 64 \), \( r = \frac{32}{64} = \frac{1}{2} \) and \( S_n = \frac{255}{2} \). The plan is to get \( (\frac{1}{2})^n \) on its own, then write the other side as a power of \( \frac{1}{2} \) as well.

Step 1: Since \( r < 1 \), putting the values into \( S_n = \frac{a(1 - r^n)}{1 - r} \):

\[ \frac{255}{2} = \frac{64\{1 - (\frac{1}{2})^n\}}{1 - \frac{1}{2}} \]

Step 2: Dividing by \( \frac{1}{2} \) is multiplying by 2, so the right side becomes:

\[ \frac{255}{2} = 128\left\{1 - \left(\frac{1}{2}\right)^n\right\} \]

Step 3: Dividing both sides by 128:

\[ \frac{255}{256} = 1 - \left(\frac{1}{2}\right)^n \]

Step 4: Getting the power on its own:

\[ \left(\frac{1}{2}\right)^n = 1 - \frac{255}{256} = \frac{1}{256} \]

Step 5: Writing 256 as \( 2^8 \):

\[ \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^8 \]

Step 6: The bases are equal, so the powers are equal:

\[ n = 8 \]

So 8 terms add up to \( \frac{255}{2} \). The whole method rests on writing both sides as powers of the same base.

Example: Hari borrowed Rs. 19,682 from Ram Naresh and agreed to pay it back in 9 instalments. Each instalment is three times the one before it. Find the difference between the first and the last instalments.

An instalment is one part of a loan paid back at a time. Because each payment is 3 times the previous one, the nine payments form a geometric series with \( r = 3 \) and \( n = 9 \). Together they must pay off the whole loan, so \( S_9 = 19682 \). The first instalment \( a \) is unknown.

Step 1: Since \( r > 1 \), putting the values into \( S_n = \frac{a(r^n - 1)}{r - 1} \):

\[ 19682 = \frac{a(3^9 - 1)}{3 - 1} \]

Step 2: Working out \( 3^9 = 19683 \):

\[ 19682 = \frac{a \times 19682}{2} \]

Step 3: Multiplying both sides by 2 and dividing by 19682 gives the first instalment:

\[ a = 2 \]

Step 4: Finding the last, or ninth, instalment with \( t_n = ar^{n-1} \):

\[ t_9 = 2 \times 3^8 = 2 \times 6561 = 13122 \]

Step 5: Subtracting the first instalment from the last:

\[ 13122 - 2 = 13120 \]

The first instalment is Rs. 2, the last is Rs. 13,122, and the difference between them is Rs. 13,120.