Arithmetic Progression

In an arithmetic progression, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_{13}-a_4=(13-4)d\), so \(77-23=9d\). Hence, \(54=9d\) and \(d=6\). If 5 were used, the difference across 9 positions would be 45, not the given difference of 54. Exam tip: subtract the term positions carefully as well as the terms.

The arithmetic mean of two numbers is their average: \((12+20)/2=16\). Thus the terms 12, 16, 20 have equal common differences of 4. Exam tip: for one middle AP term, take the average of the end terms.

In an AP, \(t_n=a+(n-1)d\). Adding the terms equally distant from \(t_n\) gives \(2[a+(n-1)d]=2t_n\). Option B is a property of a GP, not generally an AP. Exam tip: the middle term is the average of equidistant terms.

Use aₙ = a₁ + (n − 1)d. The difference between the tenth and third terms covers seven equal steps, so a₁₀ − a₃ = 7d. Hence 62 − 13 = 49 = 7d, giving d = 7. Now the third-term equation is 13 = a₁ + 2d = a₁ + 14. Therefore a₁ = 13 − 14 = −1, so option A is correct. Checking the sequence with a₁ = −1 and d = 7 gives −1, 6, 13, … and the tenth term is −1 + 9 × 7 = 62. Options B, C, and D fail to reproduce at least one of the two given terms.

Given \(a_n=5n+6\), \(a_5=5(5)+6=31\) and \(a_9=5(9)+6=51\). Therefore, \(a_5+a_9=31+51=82\). Option C, 86, is incorrect because direct substitution of the two required values of \(n\) gives 82. Exam tip: find each requested term separately by substituting its value of \(n\) before adding.

Here, the first term is 24 and the common difference is 6. Thus, \(a_n=24+(n-1)\times6\). From \(24+6(n-1)=102\), we get \(6(n-1)=78\), so \(n-1=13\) and hence \(n=14\). Option 13 is the value of \(n-1\), not the term number. Exam tip: after finding \(n-1\), remember to add 1 to obtain \(n\).

Rearranging \(2a_n=a_{n-1}+a_{n+1}\) gives \(a_n-a_{n-1}=a_{n+1}-a_n\), so consecutive differences are constant and the sequence is an AP. Option B is a GP property. Exam tip: test whether the middle term is the average of its neighbours.

Here, the first term is \(a=10\), the common difference is \(d=5\), and \(n=24\). The 24th term is \(a+(n-1)d=10+23\times5=125\). Therefore, \(S_{24}=\frac{24}{2}(10+125)=12\times135=1620\). Hence, option B is correct. An answer such as \(1632\) can result from using an incorrect last term or number of terms. Exam tip: find the \(n\)th term first and remember to use \((n-1)\).

In an arithmetic progression, the nth term is \(a_n=a+(n-1)d\). Therefore, \(a_7=a+6d\). Substituting the given values gives \(49=13+6d\), so \(36=6d\) and \(d=6\). If 5 were chosen, the seventh term would be \(13+6\times5=43\), not 49. Exam tip: the coefficient of \(d\) in \(a_n\) is always \(n-1\).

Here, the first term is \(a=18\) and the common difference is \(d=14-18=-4\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(18+(n-1)(-4)=-10\), giving \(-4(n-1)=-28\), so \(n-1=7\) and \(n=8\). Therefore, \(-10\) is the eighth term. The closest distractor, the seventh term, is incorrect because the seventh term is \(-6\). Exam tip: In a decreasing AP, remember to use a negative common difference.

(a_6=44) and (a_7=51), so the sum is (95). In exams, find both terms separately and add them.

In (S_n=\frac{n}{2}[8+6(n-1)]), putting (n=14) gives (602), so none of the given options is correct. In exams, you can check quickly by substituting options.

For an arithmetic progression, \(a_{14}-a_6=(14-6)d\). Thus, \(61-21=8d\), so \(40=8d\) and hence \(d=5\). If \(d=4\), the difference across 8 term positions would be \(32\), not \(40\). Exam tip: When subtracting two AP terms, also subtract their term numbers.

The governing concept is the general term of an arithmetic progression: aₙ = a₁ + (n − 1)d. For the fourth term, a₄ = a₁ + 3d. Substituting the given values gives 17 = a₁ + 3(4) = a₁ + 12. Therefore, a₁ = 17 − 12 = 5, so option A is correct. This can also be checked by moving backward through the progression: 17 is the fourth term, so the third term is 13, the second is 9, and the first is 5. Options B, C, and D result from subtracting the wrong number of common differences or using an incorrect difference.

The nth term is \(a_n=8n-5\). To find the position of 83, set \(8n-5=83\). This gives \(8n=88\), so \(n=11\). Hence, 83 is the 11th term. The 10th term is \(75\), so it is a close but incorrect option. Exam tip: to find the position of a given term, equate \(a_n\) to that value and solve for \(n\).

For three consecutive terms, (2(2x+1)=(x-3)+(4x-1)), giving (x=6). The middle term is the average of the two surrounding terms.

In an AP, a fixed common difference d is added to each term, so aₙ₊₁−aₙ=d. A constant ratio identifies a geometric progression, not necessarily an AP. Exam tip: compare consecutive differences to identify an AP.

In (5,11,17,23,29,\ldots), the fifth term is (29) and the common difference is (6). Check each option up to the required term.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(73=7+(12-1)d=7+11d\). Hence \(11d=66\), so \(d=6\). There are 11 gaps from the 1st term to the 12th term, not 12. Exam tip: always use \(n-1\) in the formula for the \(n\)th term.

The terms are (25,19,13,7,1,-5,\ldots), so the first negative term is the (6)th. Check terms around zero in order.