Arithmetic Progression

The direct answer is B: 100. The first term is a=30 and the common difference is d=-5 , because each term is 5 less than the preceding one. The first five terms are therefore 30, 25, 20, 15, and 10. Add them step by step: 30+25=55, 55+20=75, 75+15=90, and 90+10=100. The AP formula confirms this: S_n=\frac{n}{2}[2a+(n-1)d]=\frac{5}{2}[2(30)+4(-5)]=\frac{5}{2}(40)=100 . Option A, 90, is only the total after the first four listed terms, so it omits 10. Option B is correct. Option C, 110, adds too much and does not match the formula. Option D, 120, is also incorrect; it may come from treating the sequence as increasing or misusing the terms. A negative common difference is perfectly acceptable. Exam cue: list exactly five terms, including the fifth term 10, before adding.

From (a_6) to (a_3), move back (3) gaps, so (a_3=41-3(7)=20). When moving backward subtract the common difference.

The first term is \(a=24\) and the common difference is \(d=-6\). Using \(a_n=a+(n-1)d\), we get \(0=24+(n-1)(-6)\), so \(n=5\). Hence, 0 is the fifth term. The fourth term is 6, so it is not correct. Exam tip: To find a term’s position, substitute the given value for \(a_n\) and solve for \(n\).

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Substituting \(a=17\), \(d=0\), and \(n=10\) gives \(a_{10}=17+9\times0=17\). Hence, every term is 17. The value 170 comes from multiplying the first term by 10, which is not the rule for finding the nth term. Exam tip: when \(d=0\), every term of an AP equals its first term.

There are (6) gaps between (a_{11}) and (a_5), and (d=6), so the difference is (36). In an arithmetic progression term difference depends on the difference of positions.

The rule is \(a_n=10-3n\). Substituting \(n=1,2,3,4\) gives \(a_1=7\), \(a_2=4\), \(a_3=1\), and \(a_4=-2\), respectively. Hence, the required terms are \((7, 4, 1, -2)\). Option B, \((10,7,4,1)\), would result if counting started from \(n=0\), whereas the first term is normally obtained using \(n=1\). Exam tip: always check the starting value of the index before listing terms.

An arithmetic progression has a constant difference between consecutive terms. Here, each term increases by 6: from 10 to 16, from 16 to 22, and so on. Therefore, the first six terms are 10, 16, 22, 28, 34, and 40. The required result is the total of these six terms, not merely the sixth term. The first term is 10, the common difference is 6, and the number of terms is 6.

Using the sum formula for an arithmetic progression, \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_6=\frac{6}{2}[2(10)+(6-1)(6)]\). Thus, \(S_6=3[20+30]=3\times50=150\). Direct addition also confirms this: \(10+16+22+28+34+40=150\). Hence, option C is correct. A value such as 40 is only the sixth term, not the sum of all six terms.

From (a_2) to (a_8), the increase over (6) gaps is (30), so (d=5), hence (a_5=9+3(5)=24). Find the common difference first.

The direct answer is B: 18 and 22. First identify the common difference: 10-6=4 and 14-10=4, so every next term is found by adding 4. Starting from the last given term, 14+4=18; then 18+4=22. Thus the next two terms are 18 and 22. Option A, 16 and 18, is wrong because it adds only 2 after 14 and does not preserve the common difference. Option B is correct because both new terms follow the constant difference of 4. Option C, 20 and 24, skips the required immediate next term 18 and effectively adds 6 each time from 14. Option D, 22 and 26, skips both immediate next terms and starts too far ahead. The important idea is that “next two” means the first two terms after the last term shown, not any later pair. Memory cue: calculate d from the given terms, then add d once and again.

There are (4) gaps between (a_6) and (a_2), so the difference is (4\times11=44). For term difference look at the difference of positions.

For this arithmetic progression, the first term is \(a=27\) and the common difference is \(d=30-27=3\). Using \(a_n=a+(n-1)d\), we get \(a_n=27+(n-1)\times3=3n+24\). Hence, option A is correct. In option C, putting \(n=1\) gives 30, not the first term 27. Exam tip: always test a general term with \(n=1\) to verify the first term.

(a_7=2) and (a_8=-3), so the first negative term is the (8)th. Check carefully for the first term that goes below zero.

The governing concept is the common difference of an arithmetic progression. It is the constant amount found by subtracting one term from the next. Compute the differences: 13 − 6 = 7, 20 − 13 = 7, and 27 − 20 = 7. Because all checked consecutive differences are equal, the sequence follows an arithmetic pattern and its common difference is 7. Therefore, option C is correct. Option D, 13, is the second term rather than the change between terms. Options A and B do not equal the difference between any consecutive pair in the sequence. Checking several pairs is important because it confirms that 7 is a constant difference, not merely a result from one isolated subtraction.

In an AP, the difference between every pair of consecutive terms must be the same. Here \(7-3=4\), \(11-7=4\), but \(16-11=5\). Increasing terms alone do not make an AP. Exam tip: check consecutive differences first.

The first term is (42) and the difference is (-6), so (a_n=42+(n-1)(-6)=48-6n). In a decreasing progression, the common difference is negative.

In an arithmetic progression, the difference between consecutive terms must be constant. Here, 5−2=3, 9−5=4 and 14−9=5, which are unequal. Increasing terms alone do not form an AP. In exams, check consecutive differences first.

From (3+(n-1)8=67), (8(n-1)=64) and (n=9). To find the position, set the given term equal to (a_n).

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_6=17+(6-1)d\), so \(47=17+5d\). Hence, \(5d=30\) and \(d=6\). There are 5 gaps between the first and sixth terms, not 6. Exam tip: always use \((n-1)d\) in the formula for \(a_n\).

The governing concept is the nth-term rule of an arithmetic progression. In general, aₙ = a₁ + (n − 1)d, which can be rewritten as aₙ = (a₁ − d) + dn. Thus, in a linear expression for the nth term, the coefficient of n is the common difference. In aₙ = 55 − 5n, that coefficient is −5, so d = −5. This can be verified without relying only on the pattern: a₁ = 55 − 5 = 50 and a₂ = 55 − 10 = 45, hence a₂ − a₁ = 45 − 50 = −5. Therefore option A is correct. Option B loses the negative sign, option C is the first term, and option D is merely the constant part of the given expression, not the common difference.

For an arithmetic progression, the fourth term is \(a_4=a+3d\). Here, \(36=21+3d\), so \(3d=15\) and \(d=5\). Therefore, the correct answer is 5. Choosing 3 would be incorrect: there are three gaps from the first to the fourth term, and dividing 15 by 3 gives 5. Exam tip: use \(a_n=a+(n-1)d\) and carefully count the number of gaps between terms.

The first term is \(a=4\), and the common difference is \(d=13-4=9\). The \(n\)th term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_{10}=4+(10-1)\times9=4+81=85\). Hence, option C is correct. The value \(76\) is \(4+8\times9\), which is the 9th term, not the 10th term. Exam tip: use \(n-1\) differences to find the \(n\)th term.

In an arithmetic progression, the second term is \(a_2=a_1+d\). Therefore, \(a_1=a_2-d=18-6=12\). Hence, the correct answer is \(12\). Getting \(24\) would mean adding the common difference instead of subtracting it, which is incorrect here. Exam tip: subtract the common difference when moving back to the first term.

Here (a=60) and (d=-6), so (a_{12}=60+11(-6)=-6). In a decreasing progression, terms can become negative.

In option C, the first term is 12. The differences between consecutive terms are 8-12=-4, 4-8=-4, and 0-4=-4, so it is an AP with common difference -4. Option D also has common difference -4, but its first term is 16. Exam tip: To identify an AP, check the first term and at least two consecutive differences.

To find the first term, substitute
n=1
. Thus,
a_1=6(1)-1=5
. Therefore, the first term is 5. Option 6 is only the coefficient in the formula, not the first term. Exam tip: For any sequence, put
n=1
in the general-term formula to get its first term.