If (p,,p+7,,p+14,\ldots) is an arithmetic progression, what is the common difference?
The difference between the second and first terms is ((p+7)-p=7). In exams, subtract even with algebraic terms.
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SubjectsMathematics
समांतर श्रेणी
Arithmetic Progression for Class 9 Mathematics introduces a sequence in which consecutive terms change by a constant common difference. Students learn to recognise the pattern, identify the first term and common difference, generate further terms, and use the nth-term rule to find a required term. As part of Sequences and Progressions, the topic builds clear reasoning through number patterns, tables, and simple problems, helping learners connect a general rule with specific values and explain their steps accurately.
TOPIC PRACTICE
Up to 20 questions from this page. Select your focus, then start.
The difference between the second and first terms is ((p+7)-p=7). In exams, subtract even with algebraic terms.
View question detailsConsecutive terms are equal, so the difference is (11-11=0). In exams, a constant sequence can also be an arithmetic progression.
View question detailsWhen the common difference is (0), all terms remain (15). In exams, (d=0) means a constant sequence.
View question detailsThe difference between (18) and (25) is (7), so the missing term is (18-7=11). In exams, apply the common difference backward too.
View question detailsIn the linear rule (a_n=9n-2), the coefficient of (n) is the difference (9). In exams, take (k) as (d) in (kn+c).
View question detailsThe first term is 18 and the common difference is 5. Therefore, the sixth term is \(a_6=a+(6-1)d=18+5\times5=43\). Choosing 40 results from adding the common difference only four times. Exam tip: use \(a_n=a+(n-1)d\) for the nth term of an AP.
View question detailsThe nth term of an arithmetic progression is \(a_n=a_1+(n-1)d\). Therefore, \(a_4=13+(4-1)\times 9=13+27=40\). Hence, \(40\) is correct. \(43\) would result from adding the common difference four times, but only three differences are added to reach the fourth term. Exam tip: use \(n-1\), not \(n\), when finding \(a_n\).
View question detailsIn an arithmetic progression, the difference between consecutive terms, called the common difference, is constant; for example, 3, 8, 13 has difference 5. A geometric progression has a constant ratio, not a constant difference. Exam tip: “constant difference” identifies an AP.
View question detailsThis is a sequence of multiples of (11), so the tenth term is (110). In exams, identify multiples quickly.
View question detailsGiven \(a_n=18-4n\), substitute \(n=3\): \(a_3=18-4(3)=18-12=6\). Hence, 6 is the correct option. A value such as 8 can result from an incorrect calculation of \(4\times3\). Exam tip: while finding a term, substitute the value of \(n\) carefully into the complete expression.
View question detailsAt (n=1) it gives (14), and at (n=2) it gives (10), so (a_n=18-4n). In exams, choose the rule after checking the first term.
View question detailsThe middle term is the average of (12) and (30), so (x=\frac{12+30}{2}=21). In exams, take the average for the middle term in three-term progressions.
View question detailsHere, the first term is \(a=30\) and the common difference is \(d=38-30=8\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_7=30+(7-1)\times 8=30+48=78\). Option 76 would result from adding the difference only five times, but the seventh term requires six additions of the common difference. Exam tip: always use \(n-1\) in the formula for the \(n\)th term.
View question detailsThe first term is (12), and (6) is added each time. In exams, match both the first term and the difference.
View question detailsThe nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_6=10+(6-1)\times7=10+35=45\). The value 42 is obtained by adding the common difference only four times, so it is the fifth term. Exam tip: always subtract 1 from the term number before multiplying by \(d\).
View question detailsThe 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.
View question detailsIn 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.
View question detailsThe 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.
View question detailsIn 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.
View question detailsFrom (3+(n-1)8=67), (8(n-1)=64) and (n=9). To find the position, set the given term equal to (a_n).
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