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.

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.

Here (a=16) and (d=7), so (a_{13}=16+12\times7=100). In exams, use (a+(n-1)d) for the (n)th term.

In an AP, the difference between every pair of consecutive terms is constant, so \(q-p\), \(r-q\), and \(s-r\) must be equal. \(p+s=q+r\) alone is not sufficient. Exam tip: compare all three differences.

The direct answer is A: 48. In an arithmetic progression, the difference between consecutive terms is constant. Here the difference is \\(111-120=-9\\), and \\(102-111=-9\\), so \\(d=-9\\). The formula is \\(a_n=a+(n-1)d\\). With first term \\(a=120\\) and \\(n=9\\), \\(a_9=120+(9-1)(-9)=120+8(-9)=120-72=48\\). A is correct. B, 50, does not result from subtracting 9 eight times. C, 54, is another incorrect subtraction or counting error. D, 57, is also not the ninth term; it would reflect an incorrect number of steps or difference. There are eight gaps from the first term to the ninth term, not nine gaps. Exam cue: for a decreasing AP, keep the common difference negative and use \\(n-1\\).

To find the position of 107, put a_n = 107 in the general term: 6n + 5 = 107. Thus, 6n = 102 and n = 17. Therefore, 107 is the 17th term. The 16th term is 6(16) + 5 = 101, so it is not correct. Exam tip: To find the term number for a given value, equate a_n to that value.

The nth-term formula of an arithmetic progression is \(a_n=a+(n-1)d\). Here, \(a=21\) and \(a_{18}=106\), so \(106=21+17d\). Thus, \(17d=85\), giving \(d=5\). If the common difference were 4, the eighteenth term would be \(21+17\times4=89\), not 106. Exam tip: from the first term to the nth term, the common difference is added \(n-1\) times, not n times.

The middle term is (x=\frac{14+42}{2}=28). In exams, the middle term of three AP terms is the average.

In an AP, consecutive differences are equal: \(q-p=r-q\). Rearranging gives \(2q=p+r\), so the middle term is the average of the other two. \(q^2=pr\) is the condition for a GP. Exam tip: check the middle term.

The rule (a_n=50-6n) gives (44) at (n=1) and (38) at (n=2). In exams, match the first term in a decreasing progression.

Use the AP formula \(a_n=a+(n-1)d\). Thus, \(a_8=15+(8-1)d=15+7d\). Since \(a_8=71\), \(71=15+7d\), so \(7d=56\) and \(d=8\). Do not choose \(7\): there are seven common differences from the first term to the eighth term. Exam tip: write \((n-1)\) explicitly before substituting values in the AP formula.

Here, the first term is \(a=5\) and the common difference is \(d=9\). The \(n\)th term is \(a_n=a+(n-1)d\). So, \(5+(n-1)\times9=95\) gives \((n-1)\times9=90\), hence \(n=11\). Therefore, 95 is the eleventh term. The tenth term is 86, so it is not correct. Exam tip: when asked for a term number, equate \(a_n\) to the given term and solve for \(n\).

For an arithmetic progression, the nth term is \(a_n=a+(n-1)d\). Thus, \(a_9=19+(9-1)(-5)=19-40=-21\). Therefore, option B is correct. The value \(-19\) would result from incorrectly using 7 in place of \(n-1\). Exam tip: when the common difference is negative, check the sign after multiplication carefully.

In an AP, consecutive differences are equal: \(q-p=r-q\). Rearranging gives \(2q=p+r\), so the middle term is the average of the outer terms. Exam tip: double the middle term to test three terms quickly.

Here, the first term is \(a=3\), the common difference is \(d=7-3=4\), and \(n=20\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), \(S_{20}=\frac{20}{2}[2(3)+19(4)]=10(82)=820\). Therefore, 820 is correct. A value such as 810 can result from using an incorrect last term or replacing \((n-1)d\) incorrectly. Exam tip: for the sum of \(n\) terms, use \((n-1)d\) in the formula.

(a_{10}-a_4=6d=42), so (d=7). In exams, divide the difference of terms by the difference of positions.