Arithmetic Progression

The governing concept is the sum of the first n terms of an arithmetic progression: Sₙ = n/2 [2a₁ + (n − 1)d]. In this sequence, a₁ = 2, d = 5 − 2 = 3, and n = 10. Thus S₁₀ = 10/2 [2(2) + 9(3)] = 5[4 + 27] = 5 × 31 = 155. The same result can be checked by finding the tenth term, a₁₀ = 2 + 9 × 3 = 29, and using S₁₀ = 10/2(2 + 29) = 5 × 31 = 155. Options A, B, and C result from using an incorrect difference, term count, or arithmetic simplification. Therefore, option D is correct.

The middle term is (x=\frac{6+18}{2}=12). In exams, the middle term of three AP terms is the average.

The governing idea is the finite-sum formula for an arithmetic progression, Sₙ = n/2 [2a₁ + (n − 1)d]. The first term is a₁ = 3, the common difference is d = 8 − 3 = 5, and n = 8. Substitution gives S₈ = 8/2 [2(3) + 7(5)] = 4[6 + 35] = 4 × 41 = 164. As a verification, the eighth term is a₈ = 3 + 7 × 5 = 38; pairing the first and last terms gives S₈ = 8/2(3 + 38) = 4 × 41 = 164. The other choices do not satisfy this calculation. Thus option C is correct.

In an arithmetic progression, the difference between consecutive terms is constant. For \(a_n=7-3n\), \(a_{n+1}-a_n=[7-3(n+1)]-(7-3n)=-3\), which is fixed. For \(n^2-3\), the difference changes. Exam tip: test \(a_{n+1}-a_n\).

The rule (a_n=29-4n) gives (25) at (n=1) and (21) at (n=2). In exams, match the first term in a decreasing progression.

The formula for an arithmetic progression is \(a_n=a+(n-1)d\). Substituting \(a_6=43\), \(a=8\), and \(n=6\) gives \(43=8+5d\). Thus, \(5d=35\), so \(d=7\). If 8 were chosen, the sixth term would be \(8+5\times8=48\), not 43. Exam tip: use \(n-1\), not \(n\), in the formula for \(a_n\).

Here, the first term is \(a=4\) and the common difference is \(d=6\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(4+(n-1)6=70\) gives \(6(n-1)=66\), hence \(n=12\). Therefore, 70 is the twelfth term. The eleventh term is \(64\), so it is not correct. Exam tip: when asked for a term number, equate \(a_n\) to the given term and solve for \(n\).

The formula for the nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Thus, \(a_9=13+(9-1)(-3)=13-24=-11\). Therefore, \(-11\) is correct. \(-9\) can result from using an incorrect number of common differences. Exam tip: when \(d\) is negative, check the sign after multiplication carefully.

If \(T_p=T_q\) and \(p\ne q\), then \(a+(p-1)d=a+(q-1)d\), giving \((p-q)d=0\). Hence \(d=0\), so every term is equal. Exam tip: note the condition that the positions are distinct.

Here, the first term is \(a=1\), the common difference is \(d=3\), and \(n=20\). The sum of the first \(n\) terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Therefore, \(S_{20}=\frac{20}{2}[2(1)+19(3)]=10(59)=590\). Hence, 590 is correct. A result such as 600 may come from using an incorrect number of common differences. Exam tip: to reach the \(n\)th term from the first term, use \(n-1\) common differences.

(a_8-a_4=4d=24), so (d=6). In exams, divide the difference of terms by the difference of positions.

In an arithmetic progression, \(a_n=a_1+(n-1)d\). Therefore, \(a_5=a_1+4d\). Substituting the given values, \(27=a_1+4\times5=a_1+20\), so \(a_1=7\). If \(a_1=9\), the fifth term would be \(29\), not \(27\). Exam tip: there are \(n-1\) common differences between the first term and the \(n\)th term.

The position difference is (10) and (d=7), so the difference is (10\times7=70). In exams, use ((m-n)d) directly for the difference of two terms.

The first six terms are (7,16,25,34,43,52), and their sum is (186). In exams, you can verify the sum by listing terms from the rule.

Here, the first term is \(a=17\) and the common difference is \(d=24-17=7\). The \(n\)th term is \(a+(n-1)d\). Thus, \(17+(n-1)\times7=80\), so \((n-1)\times7=63\) and hence \(n=10\). Therefore, 80 is the 10th term. The 9th term is 73, so it is not correct. Exam tip: To find a term number, equate the given value directly to \(a+(n-1)d\).

For an arithmetic progression, \(a_n=a+(n-1)d\). Thus, \(a_2=a+d=12\) and \(a_6=a+5d=32\). Subtracting the equations gives \(4d=20\), so \(d=5\). Substituting \(d=5\) into \(a+d=12\) gives \(a=7\). Therefore, the correct answer is 7. Option 6 may seem close, but it does not satisfy the first-term calculation from \(a+d=12\). Exam tip: when two terms are given, subtract their equations first to find \(d\).

Here, the first term is \(a=30\) and the common difference is \(d=-3\). The \(n\)th term is \(a_n=a+(n-1)d\). For the term to be zero, \(30+(n-1)(-3)=0\). This gives \(30-3n+3=0\), so \(n=11\). Hence, the eleventh term is zero. The tenth term is \(3\), not zero. Exam tip: To find a term number in an AP, substitute the given term value in \(a_n\) and solve for \(n\).

For an arithmetic progression, \(a_n=a+(n-1)d\). Hence, \(a_9=2+(9-1)d=2+8d\). Given \(a_9=50\), we get \(50=2+8d\), so \(8d=48\) and \(d=6\). If \(d=5\), the ninth term would be \(2+8\times5=42\), not 50. Exam tip: in the \(n\)th-term formula, the coefficient of \(d\) is always \(n-1\).

Here, the first term is \(a=9\), the common difference is \(d=16-9=7\), and \(n=12\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{12}=\frac{12}{2}[2(9)+11(7)]=6(18+77)=570\). Therefore, 570 is correct. A value such as 558 may result from using the wrong number of terms or misusing \((n-1)\). Exam tip: write down \(a\), \(d\), and \(n\) before substituting in the sum formula.

Use the arithmetic-progression rule aₙ = a₁ + (n − 1)d. From a₃ = a₁ + 2d = 15 and a₇ = a₁ + 6d = 35, subtract the first equation from the second: 4d = 20, so d = 5. Substituting this into a₁ + 2d = 15 gives a₁ + 10 = 15, hence a₁ = 5. Another way is to note that four equal steps carry the sequence from the third to the seventh term, so each step increases by 20/4 = 5. Options A, C, and D would not produce both given terms with the same common difference. Therefore, option B is correct.