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

The seventeenth term is (105), so (S_{17}=\frac{17}{2}(9+105)=969). Finding the last term first makes calculation easier.

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.

Here (a=-8), (d=7), and (n=16), so (S_{16}=712). Do not make a sign error with a negative first term.

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.