The sum of the first n terms of an arithmetic progression is Sn = n/2 (2a + (n − 1)d), or equivalently n/2 (a + l) where l is the last term. The formula is easy. The marks are usually lost on counting n correctly.
This lesson follows finding terms from two conditions and sits inside arithmetic and geometric series.
Where does the formula come from?
Write the sum forwards and backwards. Each pair, first with last, second with second-last, adds to the same total a + l. There are n such pairs across the two rows, which gives 2S = n(a + l).
So the sum is the number of terms multiplied by the average of the first and last term. This is also a quick estimate: the average of 4 and 100 is 52, so 33 terms add to roughly 33 × 52.
How do you find the sum, step by step?
- Identify a and d from the first terms. The difference is any term minus the one before it.
- Find n. If the question gives the last term, solve the nth term rule for n. The answer must be a whole number.
- Choose a form. Use n/2 (2a + (n − 1)d) if you know d, or n/2 (a + l) if you know the last term.
- Calculate, then check with the other form.
Worked example
Find the sum 4 + 7 + 10 + … + 100.
Step 1, identify: a = 4 and d = 7 − 4 = 3.
Step 2, find n: 4 + 3(n − 1) = 100, so 3(n − 1) = 96 and n − 1 = 32. Hence n = 33.
Step 3, sum: S = 33/2 × (4 + 100) = 33/2 × 104 = 33 × 52 = 1716.
Step 4, check with the other form: 33/2 × (2(4) + 32(3)) = 33/2 × 104, which is the same, 1716.
The answer is 1716.
The mistake to watch for
The usual slip is to count the number of steps and call it the number of terms.
Mistaken working: n = (100 − 4) ÷ 3 = 32, so S = 32/2 × (4 + 100) = 16 × 104 = 1664.
The 96 ÷ 3 counts the gaps between terms. There is one more term than there are gaps, like fence posts and fence panels.
The correction is to add one: n = 33 and S = 1716. A good habit is a quick sanity check on a short list. For 4, 7, 10 the last term is 10, the gaps are (10 − 4) ÷ 3 = 2, and the terms number 3.
Check yourself
Try these on paper, then open each answer.
1. Find the sum of the first 15 terms of the arithmetic progression with a = 2 and d = 3.
Show answer
S = 15/2 × (2(2) + 14(3)) = 15/2 × 46 = 15 × 23 = 345.
Check: last term = 2 + 42 = 44, and 15/2 × (2 + 44) = 15/2 × 46 = 345.
2. Find the sum 50 + 46 + 42 + … + 10.
Show answer
a = 50 and d = −4. Solve 50 − 4(n − 1) = 10, so n − 1 = 10 and n = 11.
S = 11/2 × (50 + 10) = 11/2 × 60 = 330. Check: the average term is 30 and there are 11 terms, so 11 × 30 = 330.
3. Find the sum of all the multiples of 3 from 3 to 99.
Show answer
a = 3, d = 3 and last term 99. Then 3n = 99 gives n = 33.
S = 33/2 × (3 + 99) = 33/2 × 102 = 33 × 51 = 1683. Check: 3 × (1 + 2 + … + 33) = 3 × 561 = 1683.
Where this leads next
Next, work with ratios instead of differences in recovering a common ratio from terms. Both ideas come together later in a combined term-and-sum problem. The sequence and series laboratory shows the terms and the running total for any a, d and n you choose.
If your totals are usually right but occasionally off by one term, that is worth a few minutes of individual attention, which our teachers give in online one-to-one Additional Mathematics tuition.