Arithmetic Progression

Here, the first term is 5 and the common difference is 4. Using \(a_n=a+(n-1)d\), we get \(41=5+(n-1)\times4\). Thus, \(n-1=9\), so \(n=10\). Option 9 is incorrect because 37 is the ninth term. Exam tip: To find the number of terms up to a given last term, equate that term to \(a_n\) and solve for \(n\).

Here, \(S_4\) denotes the sum of the first four terms. Thus, \(S_4=3+8+13+18=42\). Therefore, 42 is correct. Getting 40 usually indicates an addition error in one of the terms. Exam tip: for a few terms, add directly; alternatively, use \(S_n=\frac{n}{2}[2a+(n-1)d]\).

Here, the first term is \(a=12\) and the common difference is \(d=18-12=6\). Using \(a_n=a+(n-1)d\), we get \(12+(n-1)\times6=96\). Thus, \((n-1)\times6=84\), so \(n-1=14\) and \(n=15\). Hence, 96 is the 15th term. The 14th term is \(90\), so that nearby option is incorrect. Exam tip: identify \(a\) and \(d\) first, then substitute them in \(a_n=a+(n-1)d\).

The nth term of an arithmetic progression is \(a_n=a_1+(n-1)d\). Thus, \(a_{12}=20+(12-1)(-2)=20-22=-2\). Therefore, \(-2\) is correct. The value \(0\) would be obtained for the 10th term, not the 12th term. Exam tip: retain the negative sign of \(d\) carefully in a decreasing AP.

Apply the AP sum formula Sₙ = n/2 [2a₁ + (n − 1)d]. Here a₁ = 2, d = 7 − 2 = 5, and n = 15. Therefore S₁₅ = 15/2 [2(2) + 14(5)] = 15/2 [4 + 70] = 15/2 × 74 = 15 × 37 = 555. A second check gives the fifteenth term a₁₅ = 2 + 14 × 5 = 72, so S₁₅ = 15/2(2 + 72) = 555. The distractors arise from an incorrect term count or arithmetic simplification. Hence option B is correct.

There are (6) differences from the fourth to the tenth term, so (a_{10}=26+6\times6=62). In exams, count the difference in positions.

Here, the first term is \(a=100\) and the common difference is \(d=92-100=-8\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(20=100+(n-1)(-8)\), giving \(8(n-1)=80\) and hence \(n=11\). Therefore, 20 is the 11th term. The 10th term is 28, so it is a close but incorrect option. Exam tip: keep the common difference negative for a decreasing arithmetic progression.

From (114=\frac{6}{2}[8+5d]), (d=6). In exams, keep (2a) and ((n-1)d) correct in the (S_n) formula.

Here, the first term is \(a=14\) and the common difference is \(d=5\). Using \(a_n=a+(n-1)d\), we get \(a_9=14+8\times5=54\) and \(a_{11}=14+10\times5=64\). Therefore, \(a_9+a_{11}=54+64=118\). Option 124 may result from incorrectly counting the number of differences for \(a_{11}\). Exam tip: always use \((n-1)d\) for the \(n\)th term.

Given a_n=50-3n and a_n=5, we get 50-3n=5. Thus, 3n=45 and n=15. Therefore, the 15th term is equal to 5. For the 14th term, the value is 50-3(14)=8, so it is not correct. Exam tip: To find a term number, equate the given term value to a_n and solve for n.

This is the sum of the first (9) multiples of (7), which is (315). In exams, identify sequences of multiples quickly.

In an AP, the difference between two terms equals the difference in their positions multiplied by the common difference: \(a_{12}-a_5=(12-5)d\). Thus, \(72-30=7d\), so \(42=7d\) and \(d=6\). Choosing 7 would confuse the difference in term positions with the common difference. Exam tip: when using \(a_n=a+(n-1)d\), track the term numbers carefully.

For \(a_n=7-3n\), \(a_{n+1}-a_n=[7-3(n+1)]-(7-3n)=-3\), which is constant for every \(n\). Hence it is an AP. In \(n^2-3\), the difference changes. Exam tip: always test consecutive-term differences.

Using \(T_n=a+(n-1)d\), we get \(T_p+T_r=2a+(p+r-2)d\). Since \(p+r=2q\), this becomes \(2[a+(q-1)d]=2T_q\). The product relation is not generally true. Exam tip: terms equally spaced in an AP have the middle term as their average.

Use aₙ = a₁ + (n − 1)d. The two given values produce a₃ = a₁ + 2d = 11 and a₉ = a₁ + 8d = 47. Subtracting gives 6d = 36, so d = 6. Substitution into the first equation gives a₁ + 12 = 11, hence a₁ = −1. Checking the result, the progression begins −1, 5, 11, … and after six intervals of size 6 reaches a₉ = −1 + 8 × 6 = 47. Options B, C, and D do not satisfy both equations with d = 6. Therefore, option A is correct.

Given \(a_n=6n+1\), \(a_4=6(4)+1=25\) and \(a_8=6(8)+1=49\). Therefore, \(a_4+a_8=25+49=74\). Option 72 may result from ignoring the constant term \(+1\), which is incorrect. Exam tip: substitute the correct value of \(n\) each time you find a term.

Here, the first term is 22 and the common difference is 5. Thus, the nth term is \(a_n=22+(n-1)\times5\). Substituting \(a_n=87\), we get \(22+5(n-1)=87\), so \(5(n-1)=65\), \(n-1=13\), and \(n=14\). Therefore, 87 is the 14th term of the AP. If 13 is chosen, the term would be \(82\), not 87. Exam tip: while finding a term number, handle the \(n-1\) carefully.

Write the terms as \(p, p+d, p+2d, p+3d\). Then \(p+s=p+(p+3d)=2p+3d\), while \(q+r=(p+d)+(p+2d)=2p+3d\). Hence A is correct; \(ps=qr\) is not generally true. Exam tip: the sum of outer terms equals the sum of inner terms.

Here, the first term is \(a=8\), the common difference is \(d=12-8=4\), and \(n=25\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{25}=\frac{25}{2}[2(8)+24(4)]=\frac{25}{2}(112)=1400\). Therefore, 1400 is correct. An option such as 1350 may result from an error in finding the last term or in using \((n-1)\). Exam tip: for \(n\) terms, always use \((n-1)\) with the common difference in the sum formula.

The formula for the nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_6=a+5d\). Substituting the given values gives \(41=11+5d\), so \(5d=30\) and \(d=6\). If 5 were chosen, the sixth term would be \(11+5\times5=36\), not 41. Exam tip: the coefficient of \(d\) in \(a_n\) is always \(n-1\), not \(n\).