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

This is the AP (45,42,39,\ldots), and (S_{12}=6[90+11(-3)]=342). Exam tip: treat the decrease as a negative common difference.

Putting the ratio in the sum formula gives (d=5). Exam tip: cross-multiply the ratio to form an equation.

The equation (\frac{n}{2}[4+6(n-1)]=12[2+6(n-1)]) gives (n=23). Exam tip: write (S_n) and (t_n) separately.

The numbers are (12,17,\ldots,97), and there are (18) terms. Exam tip: choose the first term of the remainder AP correctly.

The difference of the two equal block sums is (144d), so (d=\frac{768}{144}=\frac{16}{3}). Exam tip: recheck block-sum formulas carefully.

(t_6=50) and (t_{25}=126), so the sum is (\frac{20}{2}(50+126)=1760). Exam tip: count the selected terms correctly.

(S_1=6), (S_2=6), and the sum decreases afterward, so the maximum is (6). Exam tip: check small integer values.

For an AP with an odd number of terms, the sum is the product of the number of terms and the middle term. Exam tip: remember the middle-term property.

Using \(S_n=\frac{n}{2}(a_1+a_n)\), and since \(n>0\) and \(S_n=0\), we get \(a_1+a_n=0\). Hence, \(a_n=-a_1\). Neither \(d=0\) nor an even \(n\) is necessary. Exam tip: for a zero sum, check whether the first and last terms are opposites.

The numbers are (12,24,\ldots,120), and there are (10) terms. Exam tip: divisible by both means use the least common multiple.

Substituting (n=14) in (S_n=\frac{n}{2}[32+7(n-1)]) gives (1085). Exam tip: options can be checked quickly too.

The equation gives (d=\frac{18}{n+1}), so no fixed value is determined. Exam tip: check whether a variable remains in the relation.

In (11) terms, the (6)th term is the middle term, so (t_6=\frac{385}{11}=35). Exam tip: use the middle term for an odd number of terms.

Solving the two sum equations gives (a=-2) and (d=4). Exam tip: find (a) and (d) together from two sums.

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 sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Here, \(a=15\), \(d=6\), and \(n=28\). Thus, \(S_{28}=\frac{28}{2}[2(15)+27(6)]=14(30+162)=14\times192=2688\). Therefore, option C is correct. Using \(28\) instead of \(27\) would be incorrect because the common difference is added only \(n-1\) times. Exam tip: write \(n-1\) separately before substituting values in the formula.

The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Substituting the values gives \(1125=\frac{n}{2}[40+5(n-1)]\), so \(n(n+7)=450\). Hence, \(n=18\). Although \(n=-25\) is another algebraic root, the number of terms cannot be negative. Exam tip: for AP sum questions, select only a positive integral value of n.

(S_{24}=\frac{24}{2}[144+23(-4)]=624). Exam tip: handle the negative common difference carefully.

For an AP, \(a_6=a+5d=29\) and \(a_{19}=a+18d=94\). Subtracting gives \(13d=65\), so \(d=5\). Hence, \(a=29-5\times5=4\). Now, \(S_{19}=\frac{19}{2}[2a+18d]=\frac{19}{2}[8+90]=931\). Therefore, \(931\) is correct. A value such as \(950\) can result from using an incorrect first term or number of terms. Exam tip: find \(a\) and \(d\) from the given terms before applying the sum formula.

The correct answer is C, 205. For any sequence, the nth term can be obtained from consecutive partial sums: a_n = S_n − S_(n−1). Here, S_18 = 6(18²) − 5(18) = 6(324) − 90 = 1944 − 90 = 1854. Similarly, S_17 = 6(17²) − 5(17) = 6(289) − 85 = 1734 − 85 = 1649. Therefore a_18 = S_18 − S_17 = 1854 − 1649 = 205. The other options result from arithmetic or substitution errors, such as using the wrong preceding sum or mishandling the negative linear term.