Sum of Arithmetic Series

In a list of numbers such as 3, 7, 11, 15, 19, each term is found by adding the same fixed number to the term before it. Such a list is an arithmetic sequence. The fixed number is called the common difference and is written \( d \). Here \( d = 7 - 3 = 4 \). The first term is written \( a \), and the term in position \( n \) is found from this formula:

\[ t_n = a + (n - 1)d \]

When the terms of the sequence are joined with plus signs instead of commas, we get \( 3 + 7 + 11 + 15 + 19 \). This is an arithmetic series. A sequence is only a list of numbers; a series asks for their total.

The sum of the first \( n \) terms of a series is written \( S_n \). For the series above, \( S_3 = 3 + 7 + 11 = 21 \) and \( S_5 = 3 + 7 + 11 + 15 + 19 = 55 \). Adding five terms by hand is easy. Adding 50 or 500 terms one by one is slow and mistakes creep in, so we need a formula for the sum.

A shopping centre gives one of its employees a cash incentive that grows every month, based on his performance. In 2079 BS he receives Rs. 800 in Baisakh, Rs. 1000 in Jestha, Rs. 1200 in Ashadh, and so on up to Rs. 2000 in Kartik. The amount rises by Rs. 200 each month, so the amounts form an arithmetic sequence with \( a = 800 \), \( d = 200 \), \( n = 7 \) terms and last term \( t_7 = 2000 \).

The seven-month total is \( S_7 = 800 + 1000 + \dots + 2000 \). Instead of adding from the left one at a time, write the seven amounts once in order and once in reverse order side by side, then add the two numbers in each row:

Every row adds up to 2800. This is not luck. Going down the table, the number in order grows by 200 while the reversed number shrinks by 200, so their total never changes. It always equals the first term plus the last term, \( 800 + 2000 = 2800 \).

Adding all the numbers in both columns counts the series twice, so the total is \( 2S_7 \). It is made of seven equal pairs of 2800:

\[ 2S_7 = 7 \times 2800 \]

Dividing both sides by 2:

\[ S_7 = \frac{7}{2} \times 2800 = 9800 \]

So the employee receives Rs. 9800 in the seven months. Look at where the numbers came from: 7 is the number of terms and 2800 is the first term plus the last term. So the same working can be written as:

\[ S_7 = \frac{7}{2}(800 + 2000) \]

The same argument works for every arithmetic series. Take a series with first term \( a \), last term \( t_n \) and \( n \) terms. Written forwards and backwards, each of the \( n \) pairs adds up to \( a + t_n \). So \( 2S_n = n(a + t_n) \), and halving gives the first formula:

\[ S_n = \frac{n}{2}(a + t_n) \]

In many problems the last term is not given, but \( a \), \( d \) and \( n \) are known. Then we replace \( t_n \) by \( a + (n - 1)d \):

\[ S_n = \frac{n}{2}[a + \{a + (n - 1)d\}] \]

Adding the two \( a \)'s inside the bracket gives the second formula:

\[ S_n = \frac{n}{2}[2a + (n - 1)d] \]

Suppose the incentive keeps rising by Rs. 200 every month. From Baisakh 2079 to Baisakh 2080 BS there are 13 months, so Baisakh 2080 is the 13th month. How much will he get that month, and how much will he have received altogether in 13 months?

Step 1: The amount in the 13th month comes from the nth term formula with \( a = 800 \), \( d = 200 \) and \( n = 13 \):

\[ t_{13} = 800 + (13 - 1) \times 200 \]

Step 2: Working out the bracket and multiplying, since \( 12 \times 200 = 2400 \):

\[ t_{13} = 800 + 2400 = 3200 \]

So he receives Rs. 3200 in Baisakh 2080 BS.

Step 3: For the total over 13 months we know \( a \), \( d \) and \( n \), so we substitute them into the second formula:

\[ S_{13} = \frac{13}{2}[2 \times 800 + 12 \times 200] \]

Step 4: Doing the multiplications and the addition inside the bracket:

\[ S_{13} = \frac{13}{2} \times 4000 \]

Step 5: Halving 4000 and multiplying by 13:

\[ S_{13} = 13 \times 2000 = 26000 \]

So he receives Rs. 26,000 in 13 months. The first formula gives the same answer: \( \frac{13}{2}(800 + 3200) = \frac{13}{2} \times 4000 = 26000 \). Working it both ways is a good check.

When a few terms of a series are given and the sum of the first \( n \) terms is wanted, follow this order:

• Check that the series is arithmetic: neighbouring terms must differ by the same amount.
• Write down the first term \( a \), the common difference \( d \) (second term minus first term) and the number of terms \( n \).
• Choose \( \frac{n}{2}(a + t_n) \) if the last term is known, otherwise \( \frac{n}{2}[2a + (n - 1)d] \).
• Substitute, simplify inside the bracket first, then multiply by \( \frac{n}{2} \).

Worked example 1: Find the sum of the first 20 terms of the series \( 1 + 2 + 3 + \dots \).

• First term \( a = 1 \)
• Common difference \( d = 2 - 1 = 1 \)
• Number of terms \( n = 20 \)

Step 1: The last term is not given, so we put the values into \( S_n = \frac{n}{2}[2a + (n - 1)d] \):

\[ S_{20} = \frac{20}{2}[2 \times 1 + (20 - 1) \times 1] \]

Step 2: Simplifying \( \frac{20}{2} \) and the bracket:

\[ S_{20} = 10(2 + 19) \]

Step 3: Adding in the bracket and multiplying by 10:

\[ S_{20} = 10 \times 21 = 210 \]

So the sum of the first 20 natural numbers is 210.

Worked example 2: Find the sum of the first 13 terms of the series \( 2 + 4 + 6 + \dots \).

• First term \( a = 2 \)
• Common difference \( d = 4 - 2 = 2 \)
• Number of terms \( n = 13 \)

Step 1: Putting \( a \), \( d \) and \( n \) into the formula:

\[ S_{13} = \frac{13}{2}[2 \times 2 + (13 - 1) \times 2] \]

Step 2: Doing the multiplications inside the bracket:

\[ S_{13} = \frac{13}{2}(4 + 24) = \frac{13}{2} \times 28 \]

Step 3: Halving 28 and multiplying by 13:

\[ S_{13} = 13 \times 14 = 182 \]

Worked example 3: Find the sum of the first 28 terms of the series \( 5 + 17 + 29 + \dots \).

• First term \( a = 5 \)
• Common difference \( d = 17 - 5 = 12 \)
• Number of terms \( n = 28 \)

Step 1: Substituting into the formula:

\[ S_{28} = \frac{28}{2}[2 \times 5 + (28 - 1) \times 12] \]

Step 2: Since \( \frac{28}{2} = 14 \) and \( 27 \times 12 = 324 \), this becomes:

\[ S_{28} = 14(10 + 324) \]

Step 3: Adding in the bracket and multiplying by 14:

\[ S_{28} = 14 \times 334 = 4676 \]

Sometimes the sum is given and something else is missing. The sum formula is an equation linking \( S_n \), \( a \), \( d \), \( n \) and \( t_n \), so we substitute what we know and solve the equation for the unknown.

Worked example 4: An arithmetic series has first term 2, last term 29 and the sum of its first \( n \) terms is 155. Find the number of terms and the common difference.

• First term \( a = 2 \)
• Last term \( t_n = 29 \)
• Sum \( S_n = 155 \)

Step 1: The first and last terms are known, so we use \( S_n = \frac{n}{2}(a + t_n) \):

\[ 155 = \frac{n}{2}(2 + 29) \]

Step 2: Multiplying both sides by 2:

\[ 310 = 31n \]

Step 3: Dividing both sides by 31:

\[ n = \frac{310}{31} = 10 \]

Step 4: Now we put \( n = 10 \) into \( t_n = a + (n - 1)d \):

\[ 29 = 2 + (10 - 1)d \]

Step 5: Subtracting 2 from both sides:

\[ 27 = 9d \]

Step 6: Dividing both sides by 9:

\[ d = \frac{27}{9} = 3 \]

So the series has 10 terms and its common difference is 3. The series is \( 2 + 5 + 8 + \dots + 29 \).

Worked example 5: The sum of a certain number of terms of the arithmetic series \( 24 + 20 + 16 + \dots \) is 72. Find the number of terms.

• First term \( a = 24 \)
• Common difference \( d = 20 - 24 = -4 \)
• Sum \( S_n = 72 \)

Step 1: The last term is unknown, so we use the second formula:

\[ 72 = \frac{n}{2}[2 \times 24 + (n - 1)(-4)] \]

Step 2: Multiplying both sides by 2 and opening the bracket, since \( (n - 1)(-4) = -4n + 4 \):

\[ 144 = n(48 - 4n + 4) \]

Step 3: Adding the numbers in the bracket and multiplying by \( n \):

\[ 144 = 52n - 4n^2 \]

Step 4: Moving every term to one side gives a quadratic equation:

\[ 4n^2 - 52n + 144 = 0 \]

Step 5: Dividing every term by 4:

\[ n^2 - 13n + 36 = 0 \]

Step 6: Splitting the middle term, because \( -9 \) and \( -4 \) multiply to 36 and add to \( -13 \):

\[ n^2 - 9n - 4n + 36 = 0 \]

Step 7: Taking out the common factors:

\[ n(n - 9) - 4(n - 9) = 0 \]

Step 8: Taking \( (n - 9) \) out as a common factor:

\[ (n - 9)(n - 4) = 0 \]

Either \( n - 9 = 0 \), giving \( n = 9 \), or \( n - 4 = 0 \), giving \( n = 4 \). So the number of terms is 4 or 9.

Worked example 6: The third term of an arithmetic series is 0 and its tenth term is 42. Find the sum of its first 15 terms.

Here neither \( a \) nor \( d \) is known. But each known term gives one equation in \( a \) and \( d \), and two equations are enough to find both.

Step 1: From \( t_3 = a + (3 - 1)d \) and \( t_3 = 0 \), the first equation is:

\[ a + 2d = 0 \quad \text{(i)} \]

Step 2: From \( t_{10} = a + (10 - 1)d \) and \( t_{10} = 42 \), the second equation is:

\[ a + 9d = 42 \quad \text{(ii)} \]

Step 3: Subtracting equation (ii) from equation (i) removes \( a \) and leaves:

\[ -7d = -42 \]

Step 4: Dividing both sides by \( -7 \):

\[ d = 6 \]

Step 5: Putting \( d = 6 \) into equation (i) gives \( a + 12 = 0 \), so the first term is:

\[ a = -12 \]

Step 6: Now the sum formula with \( n = 15 \):

\[ S_{15} = \frac{15}{2}[2 \times (-12) + 14 \times 6] \]

Step 7: Inside the bracket \( -24 + 84 = 60 \), so:

\[ S_{15} = \frac{15}{2} \times 60 \]

Step 8: Halving 60 and multiplying by 15:

\[ S_{15} = 15 \times 30 = 450 \]

So the sum of the first 15 terms is 450.

Worked example 7: A person's monthly salary is Rs. 40,000 now. Every year a grade of Rs. 1,000 is added to his monthly salary, and the grades keep adding up year after year. What will his total income be in 6 years? How much more would he need to earn Rs. 35,00,000 in 6 years?

The grade is added to the monthly salary, but the question counts years. With 12 months in a year, each year's income is \( 1000 \times 12 \) rupees more than the year before, so the yearly incomes form an arithmetic series.

Step 1: The income of the first year, which is the first term:

\[ a = 40000 \times 12 = 480000 \]

Step 2: The rise in income each year, which is the common difference:

\[ d = 1000 \times 12 = 12000 \]

Step 3: Putting \( n = 6 \) into the second formula:

\[ S_6 = \frac{6}{2}[2a + 5d] \]

Step 4: Since \( 2a = 960000 \) and \( 5d = 60000 \):

\[ S_6 = 3[960000 + 60000] \]

Step 5: Adding in the bracket and multiplying by 3:

\[ S_6 = 3 \times 1020000 = 3060000 \]

So his total income in 6 years is Rs. 30,60,000.

Step 6: The amount still needed, found by subtracting his income from the target:

\[ 3500000 - 3060000 = 440000 \]

So he would need Rs. 4,40,000 more to earn Rs. 35,00,000 in 6 years.