Combined problems give you two separate facts, often a term and a sum, and ask for something beyond both. The method is form two equations, solve for a and d (or r), then use the result to answer the final question, including any inequality about n.
This lesson uses everything earlier in arithmetic and geometric series, especially finding terms from two conditions and the finite arithmetic sum.
How do you plan a combined question?
Before writing any algebra, list what you know and what you need. A short plan saves you from solving for unknowns you do not need.
- Underline each condition and label it term or sum.
- Write the matching equation: a + (n − 1)d for a term, n/2 (2a + (n − 1)d) for an arithmetic sum.
- Simplify both, then use the simpler one to substitute into the other.
- Solve and check both conditions.
- Use a and d to answer the last part, such as a later term or a least value of n.
Worked example
An arithmetic progression has 6th term 19 and the sum of its first 10 terms is 175. Find a and d, then find the least n for which the sum of the first n terms is greater than 1000.
Step 1, term condition: a + 5d = 19.
Step 2, sum condition: 10/2 × (2a + 9d) = 175, so 2a + 9d = 35.
Step 3, solve: from the first, a = 19 − 5d. Substitute: 2(19 − 5d) + 9d = 35, so 38 − d = 35 and d = 3. Then a = 19 − 15 = 4.
Step 4, check: the terms are 4, 7, 10, 13, 16, 19, … so the 6th is 19. The sum of ten terms is 5 × (8 + 27) = 175.
Step 5, sum in terms of n: Sn = n/2 × (8 + 3(n − 1)) = n(3n + 5)/2.
Step 6, inequality: n(3n + 5)/2 > 1000 gives 3n² + 5n − 2000 > 0. The discriminant is 25 + 24000 = 24025 = 155², so the roots are n = (−5 ± 155)/6, which are n = 25 and a negative value.
Step 7, test whole numbers: S25 = 25 × 80/2 = 1000, which is equal, not greater. S26 = 26 × 83/2 = 1079. So the least n is 26.
The answer is a = 4, d = 3 and n = 26.
The mistake to watch for
The slip is to stop at the root of the quadratic without testing the inequality.
Mistaken working: the roots are 25 and a negative number, so the answer is n = 25.
At n = 25 the sum is exactly 1000. The question asked for a sum greater than 1000, so 25 does not satisfy it.
The correction is to substitute the whole numbers either side of the root into the original sum. If the question said “at least 1000”, then n = 25 would be correct. Reading the sign in the question is part of the answer.
Check yourself
Try these on paper, then open each answer.
1. An arithmetic progression has 3rd term 10 and the sum of its first 8 terms is 104. Find a and d.
Show answer
a + 2d = 10 and 8/2 × (2a + 7d) = 104, so 2a + 7d = 26. From the first, a = 10 − 2d. Then 20 − 4d + 7d = 26, so d = 2 and a = 6.
Check: terms 6, 8, 10, … so the 3rd term is 10. S8 = 4 × (12 + 14) = 104.
2. A geometric series has first term 4 and sum to infinity 12. Find the 3rd term.
Show answer
4 ÷ (1 − r) = 12, so 1 − r = 1/3 and r = 2/3. This satisfies |r| < 1.
3rd term = 4 × (2/3)² = 4 × 4/9 = 16/9. Check: 4 ÷ (1/3) = 12.
3. An arithmetic progression has first term 5 and common difference 2. Find the least number of terms needed for the sum to exceed 500.
Show answer
Sn = n/2 × (10 + 2(n − 1)) = n(n + 4). Then n² + 4n − 500 > 0, and the positive root is about −2 + √504 ≈ 20.45.
Test: S20 = 20 × 24 = 480, which is below 500. S21 = 21 × 25 = 525, which is above. So the least n is 21.
Where this leads next
Test the whole module in the mixed practice set, which includes problems of this kind. The sequence and series laboratory helps you explore how the sum changes with n, and the non-calculator working trainer supports the arithmetic in the last steps.
If choosing which condition to use first is the hard part, our teachers can work through your own attempts in online one-to-one Additional Mathematics tuition.