Finding the sum of the first $n$ terms of an AP

This is the sum of the first (25) multiples of (6), so the total distance is (1950) meters. In sums of multiples, (a=d).

Given \(S_n=5n^2-n\), substitute \(n=9\): \(S_9=5(9)^2-9=5\times81-9=405-9=396\). Hence, 396 is the correct option. 405 is only \(5\times81\); the final subtraction of 9 must also be done. Exam tip: after substituting \(n\), evaluate the power first, then multiply and subtract.

The governing concept is the sum of the first n terms of an arithmetic progression. Here, the first term is a = 7, the common difference is d = 5, and the last term is 82. Using 82 = 7 + (n − 1)5 gives n = 16. Therefore, Sₙ = n/2(first term + last term) = 16/2 × (7 + 82) = 8 × 89 = 712. Hence option B is correct; the other values result from an incorrect term count or sum.

From (\frac{n}{2}[50-4(n-1)]=105), (n=7). Even in a decreasing AP, (n) must be a positive integer.

The governing concept is the sum of the first n terms of an AP. The common difference is constant: d = 16 − 9 = 7, so a = 9 and n = 13. Applying S_n = n/2[2a + (n − 1)d], we get S_13 = 13/2[2(9) + 12(7)] = 13/2[18 + 84] = 13/2 × 102 = 13 × 51 = 663. Therefore, option C is correct. A useful check is to find the thirteenth term: a_13 = 9 + 12 × 7 = 93. The average of the first and last terms is (9 + 93)/2 = 51, and 13 × 51 = 663. The other options arise from using an incorrect common difference, term count, or arithmetic operation.

The sum of the first \(n\) terms is \(S_n=\frac{n}{2}(a+l)\). Using \(l=a+(n-1)d\) gives option A. Option B represents only the \(n\)th term. Exam tip: distinguish the formula for a sum from that for an \(n\)th term.

Direct answer: 1121, so option C is correct. When the first term a, last term l, and number of terms n are known, use Sₙ = n/2(a + l). This formula works because an AP has equally spaced terms, so the average of all its terms is the average of the first and last terms. Here a = 14, l = 104, and n = 19. Therefore S₁₉ = 19/2(14 + 104) = 19/2×118 = 19×59 = 1121. Option C follows directly. Option A, 1101, option B, 1111, and option D, 1131, are nearby values but do not result from multiplying 19 by the correct average 59. There is no need to find the common difference or list all 19 terms. A useful check is that 14 and 104 have average (14 + 104)/2 = 59; multiplying this average by 19 gives 1121. Memory cue: AP sum equals number of terms multiplied by the average of the first and last terms.

The governing concept is subtraction of partial sums. S₁₅ represents the sum of the first 15 terms, while S₁₀ represents the sum of the first 10 terms. When the latter is subtracted from the former, the first 10 terms cancel and only terms 11 through 15 remain. Therefore, required sum = S₁₅ - S₁₀ = 465 - 240 = 225. Option D is correct. This method does not require finding the first term or the common difference. A, B and C are distractors produced by incorrect subtraction or by counting the block incorrectly. Because the question includes both boundary terms, subtracting S₁₀ is exactly appropriate; subtracting S₁₁, for example, would omit the 11th term.

The governing concept is the AP sum formula written in terms of the first and last terms: Sₙ = n/2 (a + l). The question gives a + l = 150 and Sₙ = 1800. Substitution gives 1800 = n/2 × 150 = 75n. Dividing both sides by 75 gives n = 1800/75 = 24. Thus the AP contains 24 terms, so option D is correct. The common difference is not needed because the sum of the first and last terms is already supplied. The other options would produce totals of 1500, 1650 and 1950 respectively when multiplied by 75, so they do not satisfy the given total of 1800.

If \(S_n\) is the sum of the first \(n\) terms, the sum from the 21st through the 30th term is \(S_{30}-S_{20}\). The subtraction removes the first 20 terms and leaves exactly terms 21 to 30. Using the given rule, \(S_{30}=3(30)^2+4(30)=2700+120=2820\), while \(S_{20}=3(20)^2+4(20)=1200+80=1280\).

Therefore, the required sum is \(2820-1280=1540\). This matches option C. It is important not to use \(S_{30}-S_{21}\), because that would remove the first 21 terms and begin with the 22nd term. The endpoints are included: the calculation must contain both the 21st and the 30th terms, and subtracting \(S_{20}\) does exactly that.

The governing concept is the difference between two partial sums. S₁₂ is the total of terms 1 through 12, and S₆ is the total of terms 1 through 6. Subtracting S₆ from S₁₂ removes the first six terms and leaves exactly terms 7 through 12. Hence the required sum is S₁₂ - S₆ = 420 - 150 = 270. Therefore option A is correct. No common difference or individual term needs to be calculated. Options B, C and D are close numerical distractors, but each fails to equal the exact difference of the supplied partial sums. The inclusion of the 7th term is why S₆, rather than S₇, must be subtracted.

Direct answer: 760, so option B is correct. The first term is a = 95, and the common difference is d = 89 − 95 = −6. There are n = 20 terms. First find the last term using aₙ = a + (n − 1)d: a₂₀ = 95 + 19(−6) = 95 − 114 = −19. Now use Sₙ = n/2(a + l), where l is the last term. Thus S₂₀ = 20/2[95 + (−19)] = 10×76 = 760. Option B is therefore correct. Option A, 730, option C, 790, and option D, 820, result from an incorrect last term, an incorrect difference, or an arithmetic error. The negative common difference only tells us that the terms decrease; it does not mean the sum must be negative. Notice also that the average of the first and last terms is (95 − 19)/2 = 38, and 20×38 = 760, which confirms the result. Memory cue: for a finite AP, calculate the last term carefully before using n/2(a+l).

Direct answer: 3315, so option D is correct. Use the AP sum formula Sₙ = n/2[2a + (n − 1)d]. Substitute n = 30, a = −20, and d = 9: S₃₀ = 30/2[2(−20) + 29×9] = 15[−40 + 261] = 15×221 = 3315. Hence option D is correct. As a check, calculate the last term: a₃₀ = a + 29d = −20 + 261 = 241. Then S₃₀ = 30/2(−20 + 241) = 15×221 = 3315 again. Option A, 3225, option B, 3255, and option C, 3285, do not follow from the formula and may result from mishandling the negative first term or multiplying incorrectly. Keep the minus sign inside the bracket: 2a means 2×(−20) = −40, not +40. The sequence begins negatively but increases by 9, which explains why the total can be positive. Memory cue: substitute signs before simplifying.

The governing concept is the finite AP sum formula S_n = n/2(a + l), where a and l are the first and last terms. The question already gives their combined value, a + l = 260, and gives the total sum as S_n = 4160. Substitution gives 4160 = n/2 × 260 = 130n. Dividing both sides by 130 yields n = 4160/130 = 32. Therefore option D is correct. There is no need to determine the common difference or the individual first and last terms, because only their sum is used in this formula. Checking the distractors, 28, 30, and 34 would produce totals 3640, 3900, and 4420 respectively, not 4160.

Direct answer: 600, so option A is correct. S₁₆ means the sum of terms 1 through 16, while S₈ means the sum of terms 1 through 8. Subtracting S₈ from S₁₆ removes the first eight terms and leaves exactly terms 9 through 16: required sum = S₁₆ − S₈ = 880 − 280 = 600. Option A is therefore correct. Option B, 580, option C, 620, and option D, 640, do not equal the difference of the two given partial sums. Adding 880 and 280 would be wrong because it would count the first eight terms twice. Reversing the subtraction would produce −600 and would not represent the requested later block in this context. No value of a or d is needed, making subtraction the shortest valid method. Memory cue: the sum from rth to kth term is S_k − S_{r−1}; here it is S₁₆ − S₈.

For an arithmetic progression with n terms, the sum is S_n = n/2(a + l), where a and l are the first and last terms. The question gives a + l = 340 and S_n = 5780. Substitution gives 5780 = n/2 × 340 = 170n. Dividing by 170, n = 5780/170 = 34. Hence option B is correct. The common difference and the individual values of the first and last terms are unnecessary because their sum is already supplied. Options 32, 36, and 38 can result from forgetting the factor 1/2, doubling the wrong quantity, or making an arithmetic error while dividing. The result is also reasonable because 34 × 170 equals 5780 exactly.

The governing formula for the sum of an arithmetic progression is S_n = n/2(a + l), where a is the first term and l is the last term. The question gives the combined value a + l = 420 and the total sum S_n = 7350, so substitute these directly: 7350 = n/2 × 420 = 210n. Dividing both sides by 210 gives n = 7350/210 = 35. Therefore, option C is correct. The common difference, first term separately, and last term separately are unnecessary here because their sum is already supplied. Options A, B, and D result from an incorrect division or substitution.

Answer: C, 120. Use the AP sum formula S_n = n/2[2a + (n − 1)d]. Here n = 20, a = 25, and d = −2. Substitution gives S_20 = 20/2[2(25) + 19(−2)] = 10[50 − 38] = 10×12 = 120. Thus option C is correct. The negative common difference means that the terms decrease: 25, 23, 21, and so on. Another way to check is to find the last term: a_20 = a + 19d = 25 − 38 = −13. The average of the first and last terms is (25 + (−13))/2 = 6, and 20×6 = 120. Option A, B, and D do not agree with the formula. The main warning is not to replace 19d by 20d and not to lose the negative sign.

The governing concept is the sum of an arithmetic progression. The weekly numbers form an AP with first term a = 35, common difference d = 8, and number of terms n = 16. Use S_n = n/2[2a + (n−1)d]. Thus S_16 = 16/2[2(35) + 15(8)] = 8[70 + 120] = 8 × 190 = 1520. Therefore option B is correct. A common mistake is to calculate only the 16th week, which would be a_16 = 35 + 15(8) = 155, rather than the total for all 16 weeks. The other options do not satisfy the required sum formula and likely reflect an incorrect number of increases or an arithmetic error.

The correct answer is 225, option C. If S_n denotes the sum of the first n terms, then the nth term is obtained by subtracting two consecutive sums: a_n = S_n − S_(n−1). First calculate S_12 = 5 × 12 × (2 × 12 − 1) = 60 × 23 = 1380. Next calculate S_11 = 5 × 11 × (2 × 11 − 1) = 55 × 21 = 1155. Therefore a_12 = S_12 − S_11 = 1380 − 1155 = 225. Option A, 215, is not this difference; option B, 220, can result from mishandling the expression 2n − 1; and option D, 230, also does not equal the difference of the two required partial sums. Remember: subtracting the sum up to n − 1 from the sum up to n leaves only the nth term.

The governing idea is the paired-term property of an arithmetic progression. For an even number of terms, the sum can be written as S₂ₘ = m(tₘ + tₘ₊₁), because the first and last terms, second and second-last terms, and so on have equal pair sums. Here 20 = 2 × 10, so S₂₀ = 10(t₁₀ + t₁₁). Substituting the given values gives 1160 = 10(56 + t₁₁). Dividing by 10 gives 116 = 56 + t₁₁, hence t₁₁ = 60. Therefore option B is correct. The other options result from an incorrect pairing or arithmetic error.

The correct answer is 17, option C. In the progression, the first term is a = x, the common difference is d = 5, and the number of terms is n = 24. Use the AP sum formula S_n = n/2[2a + (n − 1)d]. Substitution gives 1788 = 24/2[2x + 23 × 5] = 12(2x + 115). Dividing by 12 gives 149 = 2x + 115. Hence 2x = 34 and x = 17. Option A, 15, option B, 16, and option D, 18, do not produce the stated total when substituted into the formula. The factor 23 is important: 24 terms contain 23 equal increases from the first term to the last term. A useful check is that the last term is 17 + 23 × 5 = 132, so the average of the first and last terms is (17 + 132)/2 = 74.5 and 24 × 74.5 = 1788.

The correct answer is 480 metres, option C. The gaps form an arithmetic progression: 18, 20, 22, ... Thus a = 18, d = 2, and n = 15. Find the fifteenth gap first: a_15 = a + (15 − 1)d = 18 + 14 × 2 = 46 metres. Now add the 15 gaps using S_n = n/2[2a + (n − 1)d]: S_15 = 15/2[2 × 18 + 14 × 2] = 15/2(36 + 28) = 15/2 × 64 = 480 metres. Another check uses the AP average: the average gap is (18 + 46)/2 = 32 metres, so the total is 15 × 32 = 480 metres. Option A, 450, option B, 465, and option D, 495, do not equal this sum. Do not confuse the fifteenth gap, 46 metres, with the total of all 15 gaps.