Q&A for How to Find the Sum of an Arithmetic Sequence

A sequence is arithmetic if there is a constant difference between any term and the terms immediately before and after it: for example, if each term is 7 more than the term before it.

You do this so that you can find the average of the two numbers. For example, if you were finding the average between 7, 12, and 8, you would add them up (27) and divide them by the number of values you have. In this case, you have three numbers, so you'd divide 27 by 3 to get an average of 9. In the case of the sum of an arithmetic sequence, you have two numbers that you are finding the average of, so you divide it by the amount of values you have, which is two.

You will find that 1 + 50 = 2 + 49 = 3 + 48 (and so on). Multiply the sum, which is 51, by half of the last term. You have the equation 51 × 25 = 1275. The sum is therefore 1275.

The sequence would be 1, 3, 5, 7, 9, etc. Since 100 is even, you would really look at the odd numbers 1-99. So the first term is 1, and the last term is 99. Since half of the numbers between 1 and 100 are odd, the number of terms in the sequence is 50. So, the average of the first and last term is 50, since (1 + 99)/2 = 50. Multiplying the average by the number of terms, you get 50 x 50 = 2500. So the sum of this sequence is 2,500.

Because the sum of an arithmetic sequence is equal to the average of the first and last terms multiplied by the number of terms.

This formula was derived many centuries ago through simple inspection of arithmetic sequences.

Yes, the sigma sign is used in the formula for the sum of an arithmetic sequence.

It depends on what other information you're given. If you know the last term of the sequence, the number of terms, and the sum of the sequence, you can use the sum formula given above to solve for the first term. If you know the sum and all the other terms, you can subtract the sum of the other terms from the sum of the total sequence to find the first term.

The sum alternates between 1 and 0 with each successive term. The sum is 1 after considering each odd-numbered term (that is, after considering the first, third, fifth, seventh, etc. term), so the sum is 1 after adding the 99th term.

If you use the formula in the article, the answer would be 5. a1, the first term, is -22 while an, the tenth term, is 23. This can be figured out because to get from each term you need to add 5. The number of terms in this case is 10. 10((-22+23)/2) = 10(1/2) = 5.

Sometimes the last term is given. If not, you would have to know which specific term the last term is (e.g., the 10th term, the 99th term, etc.) You can find a specific term in an arithmetic sequence as shown in Method 3 above.

Assuming you're asking about inserting three terms into an arithmetic sequence that starts at 11 and finishes at -7, that's a gap of 18. Inserting three terms means creating four equal gaps. That means each gap would be 18/4 or 4½. The first term would be 11 - 4½ = 6½. The second term is 6½ - 4½ = 2. The third term is 2 - 4½ = -2½. Then -2½ - 4½ = -7.

The ellipsis means "and the numbers that follow." Use the sum formula as you normally would.

Use the regular sum formula with (-3) as the common difference.

If you multiply n by d and subtract that product from "a sub n," you'll get the first term of the sequence. Then you can use the sum formula (and also list the sequence if you want to).

12+18+24+30+36+42 = 162.

The two formulas do look alike, but that's coincidental. There's really no relationship between them.

You'd have to do this by inspection, because it's not an arithmetic sequence (since there's no common difference between terms). The difference between terms increases by 2 with each term, so the next three terms are 25, 36 and 49.

You don't have to find the formula. It's given to you in the article above.

If the sequence is geometric, use the applicable formula, which you can easily find online. If the sequence is anything else, there is no formula available.

2 + 5 + 9 + 12 = 28. That sequence repeats 200 / 4 = 50 times in 200 numbers. The sum of the first 200 numbers is (28)(50) = 1400.

There's not enough information to answer that. You'd also have to know either the first term or the seventh term.

Using the formula above, n = 60. So the sum equals (60)(1 + 60) / 2 = (60)(61) / 2 = 3660 / 2 = 1,830.

Let x be the first term and n be the increment per term. Then the 5th term is x+(5-1)n and the 9th term is x+(9-1)n. This means x+4n + x+8n = 6, => 2x+12n = 6. Divide both sides by 2 => x+6n = 3. The 25th term is x+24n = -24. Substitute x = 3-6n in the previous equation, to get 3-6n+24n = -24 => 3+18n = -24 => 18n = -27. Divide both sides by 9 => 2n = -3, n = -3/2 = -1.5. Thus x is 3-6(-1.5) = 3+9 = 12. The fifteenth term is x+14n = 12+14*(-1.5) = 12-21 which is -9.

The sum is 14[(12 + 35/4) / 2] = 14[(48/4 + 35/4) / 2] = 14(83/4) / 2 = 7(83/4) = 581/4 =145¼.

Since that is neither an arithmetic sequence nor a geometrical sequence, we have to answer that question by inspection. The first term is 1/1, the second term is 1/2, the third term is 1/3, and so on. Therefore, the tenth term is 1/10.

The easiest way to do this is first to calculate the annual income for each of the four years. In the first year it's 12 times the monthly salary of $5000, or $60,000. In the second year she made $60,000 + $500, or $60,500. In the third year it's $60,500 + $500 = $61,000. In the fourth year she made $61,500. Now use the formula for the sum of an arithmetic sequence, where n = 4, and the first and fourth terms are 60,000 and 61,500, respectively. The sum total after four years is 4[(60,000 + 61,500) / 2] = 4(121,500 / 2) = $243,000, which is her total salary for the first four years.

If a "consecutive sequence" is a series of consecutive numbers, you would subtract the first number from the last number and then add one. For example, if the first number is 11, and the last number is 35, you would subtract 11 from 35, which is 24, and then add one to make 25. That's how many numbers there are in the sequence (including the first and last numbers).

Yes. Sn = 4(10 + 22/4) = 4(40/4 + 22/4) = 4( 62/4) = 62, an even number.

Sum = const ( (first + last) / 2). Sum / const = (first + last) / 2. 2(Sum / const) = first + last. So finally, last = 2(Sum / const) - first.