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

The correct answer is option A: 15 terms. In the AP, the first term is \(a=5\) and the common difference is \(d=16-5=11\). The sum formula is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Substitute the values: \(1230=\frac{n}{2}[10+11(n-1)]\). Simplify the bracket: \(10+11n-11=11n-1\), so \(1230=\frac{n(11n-1)}{2}\). Testing the given options is quick. For \(n=15\), the last term is \(5+14(11)=159\), and the sum is \(\frac{15}{2}(5+159)=\frac{15}{2}\times164=1230\). Thus A works. For 16 terms the sum is 1392, for 17 it is 1570.5, and for 18 it is 1764, so B, C and D are wrong. A common mistake is using 16 as the difference; the difference is 11.

Putting (n=16) in (S_n=\frac{n}{2}[8+12(n-1)]) gives (1504). Checking by substituting options is safe.

The numbers are (126,135,\ldots,297), and there are (20) terms, so the sum is (4230). In boundary questions, choose the first and last terms carefully.

The numbers are (112,126,\ldots,336), and there are (17) terms, so the sum is (3808). Find the first and last suitable multiples.

The numbers are (60,75,\ldots,240), and there are (13) terms, so the sum is (1950). Make the correct sequence by checking the limits.

The numbers are (208,224,\ldots,496), and there are (19) terms, so the sum is (6688). Do not forget to find the number of terms.

The sum of the first (n) odd numbers is (n^2), so (n^2=625) and (n=25). Recognizing perfect squares is useful.

The sum of the first (n) even numbers is (n(n+1)), so (27\times28=756). Do not confuse (n) with the last even number.

The sum of the first n odd natural numbers is n². Therefore, the sum of the first 26 odd numbers is 26² and that of the first 14 odd numbers is 14². Remaining sum = 26² − 14² = 676 − 196 = 480. A value such as 468 may result from using an incorrect average for the 12 remaining terms. Exam tip: Recall directly that the sum of the first n consecutive odd numbers is n².

The sum of the first 28 even natural numbers is \(28(28+1)=812\). The sum of the first 11 even natural numbers is \(11(11+1)=132\). Therefore, the sum of the remaining numbers is \(812-132=680\). A value such as 660 can result from an error in counting terms or subtraction. Exam tip: the sum of the first \(n\) even natural numbers is \(n(n+1)\).

Sum equals average (\times) number of terms, so (64\times15=960). If the average is given, the long formula is not necessary.

The formula for the average is \(\text{Average}=\frac{\text{sum of all terms}}{\text{number of terms}}\). Therefore, \(\frac{1170}{18}=65\). Hence, 65 is the correct answer. Option 64 is incorrect because \(18\times64=1152\), not the given sum of 1170. Exam tip: For an average question, divide the total sum by the number of terms.

The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}(a+l)\). Thus, \(912=\frac{12}{2}(18+l)=6(18+l)\). Hence, \(18+l=152\), so \(l=134\). If \(l=136\), the sum would be \(6(18+136)=924\), not the given sum. Exam tip: while finding the last term, first isolate \(a+l\) in the sum formula.

For an AP, the sum of the first \(n\) terms is \(S_n=\frac{n}{2}(a+l)\). Thus, \(975=\frac{15}{2}(a+97)\). This gives \(a+97=130\), so \(a=33\). If 35 were used as the first term, the sum would not be 975. Exam tip: When the last term is given, directly use \(S_n=\frac{n}{2}(a+l)\).

Using (S_n=\frac{n}{2}(a+l)), (S_{18}=\frac{18}{2}\times144=1296). If (a+l) is directly given, use it immediately.

The numbers are (12,16,\ldots,96), and there are (22) terms, so the sum is (1188). Apply the two-digit limit carefully.

The multiples are (100,125,\ldots,275), and there are (8) terms, so the sum is (1500). Decide the first and last terms according to the limit.

The first term is (9) and the twentieth term is (104), so (S_{20}=\frac{20}{2}(9+104)=1130). Use (a_n) to find the first and last terms.

Direct answer: option C, 1035. The formula a_n=7n-9 gives each term directly. For the first term, put n=1: a1=7(1)-9=-2. For the eighteenth term, put n=18: a18=7(18)-9=126-9=117. These terms form an AP with common difference 7, so use S18=18/2(a1+a18)=9(-2+117)=9×115=1035. Option A, 995, is too small and does not result from the correct endpoints. Option B, 1015, also reflects an arithmetic or indexing error. Option C, 1035, matches the exact sum. Option D, 1055, is too large and may result from mishandling the negative first term. The key point is that a_n uses the position n: use n=1 for the first term and n=18 for the last term. As a check, direct summation of the linear rule gives the same result.

The correct answer is option D: 568. The nth term is \(a_n=3n+10\). Put \(n=1\): the first term is \(a_1=3(1)+10=13\). Put \(n=16\): the sixteenth term is \(a_{16}=3(16)+10=58\). For an AP, the sum of the first n terms is \(S_n=\frac{n}{2}(a_1+a_n)\). Therefore, \(S_{16}=\frac{16}{2}(13+58)=8\times71=568\). Option D is correct. Option A, 538, results from an incorrect addition or term value. Option B, 548, is also too small and does not come from the correct formula. Option C, 558, is close but still misses the exact total; closeness is not enough in an arithmetic question. The first and last terms must be added and multiplied by half the number of terms. A useful check is that the average term is \((13+58)/2=35.5\), and \(16\times35.5=568\).

The first term is (48) and the fourteenth term is (22), so (S_{14}=\frac{14}{2}(48+22)=490). The same sum formula works for a decreasing progression.

The twelfth deposit is (265) rupees, so the total is (S_{12}=\frac{12}{2}(45+265)=1860) rupees. Keep the unit in mind in money questions.

The tenth row has (395) chairs, so the total is (S_{10}=\frac{10}{2}(80+395)=2375). Treat rows as the number of terms.

The eighteenth row has (229) bricks, so the total is (S_{18}=\frac{18}{2}(25+229)=2286). In word problems, convert the pattern into an arithmetic progression.

The governing concept is the sum of the first n terms of an arithmetic progression. The first five terms are 2, 7, 12, 17, and 22, because the common difference is 5. Direct addition gives S₅ = 2 + 7 + 12 + 17 + 22 = 60, so option B is correct. The standard formula confirms this: Sₙ = n/2[2a₁ + (n − 1)d], hence S₅ = 5/2[2(2) + 4(5)] = 5/2(24) = 60. Option A, 55, omits 5 from the total; option C adds an extra 5; and option D is larger still because it does not represent the sum of exactly the first five terms.