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

The 12th-day saving is \(50+11\times10=160\). Hence, the sum is \(\frac{12}{2}(50+160)=1260\). Riya used \(n\) instead of \(n/2\). Exam tip: find the last term before applying the sum formula.

Direct answer: option A, 712. In the AP -8,-1,6,..., the first term is a=-8, the common difference is d=7, and n=16. The sixteenth term is a16=-8+(16-1)×7=-8+105=97. Therefore S16=16/2×(-8+97)=8×89=712. Option A, 712, is correct. Option B, 722, is ten too large and can arise from mishandling the negative first term. Option C, 732, does not follow from the AP formula and is still larger than the correct result. Option D, 742, is also incorrect; it may come from adding an extra amount or using the wrong number of steps. A useful check is that the average term is (-8+97)/2=44.5, and 16×44.5=712. The difference is positive, so terms increase, but the initial negative sign must remain.

The fifteenth term is (-36), so (S_{15}=\frac{15}{2}(20-36)=-120). In a decreasing progression, take (d) as negative.

The twenty-first term is (185), so (S_{21}=\frac{21}{2}(5+185)=1995). For larger (n), write the last term first.

The twentieth term is (128), so (S_{20}=\frac{20}{2}(14+128)=1420). Keep both the last term and the number of terms correct.

The nineteenth term is (78), so (S_{19}=\frac{19}{2}(-12+78)=627). The same formula works even when the progression starts with a negative term.

The twenty-second term is (171), so (S_{22}=\frac{22}{2}(3+171)=1914). The average of the first and last terms is useful.

The eighteenth term is (-26), so (S_{18}=\frac{18}{2}(25-26)=-9). The sum of a decreasing progression can also be negative.

This is the sum of the first (14) multiples of (11), so (11\times\frac{14\times15}{2}=1155). For multiples, use the sum of natural numbers.

The seventeenth term is (-40), so (S_{17}=\frac{17}{2}(40-40)=0). With opposite equal end terms, the sum can be zero.

The fifteenth term is (189), so (S_{15}=\frac{15}{2}(7+189)=1470). With a large difference, finding the last term is important.

The twenty-third term is (104), so (S_{23}=\frac{23}{2}(16+104)=1380). With an odd number of terms, the middle term can also check the sum.

The eighteenth term is (82), so (S_{18}=\frac{18}{2}(-20+82)=558). Keep the sign of the negative first term in mind.

The sixteenth term is (227), so (S_{16}=\frac{16}{2}(2+227)=1832). Use ((n-1)d) while finding the last term.

For an arithmetic progression, the terms increase or decrease by a constant difference. When the first term, last term, and number of terms are known, the sum can be found by pairing terms: the first and last have the same total as the second and second-last, and so on. This gives the efficient formula \(S_n=\frac{n}{2}(a+l)\), where \(a\) is the first term and \(l\) is the last term.

Here, \(n=19\), \(a=18\), and \(l=126\). Substitution gives \(S_{19}=\frac{19}{2}(18+126)=\frac{19}{2}\times144=19\times72=1368\). Since there are an odd number of terms, the pairing idea still works with the middle term left unpaired. Thus the required sum is 1368, so option B is correct. The other values result from an arithmetic or substitution error.

The sum of the first n terms of an AP is \(S_n=\frac{n}{2}(a+l)\), where \(a\) is the first term and \(l\) is the last term. Thus, \(S_{24}=\frac{24}{2}(9+147)=12\times156=1872\). Therefore, 1872 is the correct answer. An option such as 1852 results from an arithmetic error in addition or multiplication. Exam tip: When the first and last terms are given, use \(\frac{n}{2}(a+l)\) directly.

In an AP, the average of the first and last terms is \(\frac{a+l}{2}\). Multiplying this by the number of terms \(n\) gives \(S_n=\frac{n}{2}(a+l)\). Option D represents a difference, not a sum. Exam tip: use average × number of terms.

The correct formula is \(S_n=\frac{n}{2}(a+l)\), because \(\frac{a+l}{2}\) is the average of the first and last terms, multiplied by n terms. Option C gives only the average, not the sum. Exam tip: check the factor n.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\), where \(a\) is the first term and \(l\) is the last term. Thus, \(S_{18}=\frac{18}{2}(85+0)=9\times85=765\). Therefore, 765 is correct. A last term of 0 must still be included in the formula; it should not be omitted. Exam tip: When the first term, last term, and number of terms are given, use \(S_n=\frac{n}{2}(a+l)\) directly.

The sum of the (9)th to (17)th terms is (S_{17}-S_8=792). For a group of consecutive terms, take the difference of partial sums.

The governing concept is the meaning of a partial sum. S₁₄ is the sum of the first fourteen terms, and S₆ is the sum of the first six terms. Subtracting the smaller partial sum from the larger one cancels the first six terms and leaves the terms numbered 7 through 14. Therefore the required sum is S₁₄−S₆=665−165=500. Hence option A is correct. There is no need to find the first term or common difference, because the requested consecutive block is obtained directly from the two given sums. The alternative values 510, 520, and 530 do not follow from the stated subtraction and most likely arise from a simple arithmetic error. This method is valid because both partial sums refer to the same arithmetic progression and begin at its first term.

The sum of the (6)th to (12)th terms is (S_{12}-S_5=378). The sum of middle terms is found from the difference of partial sums.

The direct answer is option B, 1298. S_9 is the sum of the first nine terms, while S_20 is the sum of the first twenty terms. To obtain the sum from the 10th through the 20th terms, subtract the first nine terms from the first twenty terms: S_20 − S_9 = 1640 − 342 = 1298. Option A, 1278, results from an arithmetic subtraction error. Option B is correct because the first nine terms are removed, leaving terms 10 to 20, exactly the requested eleven terms. Option C, 1318, and option D, 1338, do not equal the required difference. It is not necessary to find the first term or common difference. Memory cue: the sum from term k+1 through term m is S_m − S_k; check the starting index carefully.

Direct answer: option C, 15 terms. The AP is 6,14,22,..., so a=6 and d=8. Its sum is S_n=n/2[2a+(n-1)d]=n/2[12+8(n-1)]. Simplify the bracket: 12+8n-8=8n+4, so S_n=n(4n+2)=4n^2+2n. Set this equal to 930: 4n^2+2n=930, hence 2n^2+n-465=0. Testing the answer choices is quick: for n=15, S15=15/2[12+8×14]=15/2×124=15×62=930. Thus option C is correct. Option A, 13, gives 4(169)+26=702. Option B, 14, gives 4(196)+28=812. Option D, 16, gives 4(256)+32=1056. Therefore only 15 produces 930. Do not confuse the common difference 8 with the first term or omit the n-1 factor.

Putting (n=16) in (S_n=\frac{n}{2}[20+9(n-1)]) gives (1240). The option method is fast when finding (n).