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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 5 questions from this page. Select your focus, then start.
In an arithmetic progression, \(a_n=a+(n-1)d\). Therefore, \(a_8=a+7d\). Substituting the given values gives \(66=17+7d\), so \(7d=49\) and \(d=7\). Hence, option C is correct. If \(d=8\), the eighth term would be \(17+7\times8=73\), not 66. Exam tip: use \(n-1\), not \(n\), in the formula for the \(n\)th term.
Here, the first term is \(a=24\) and the common difference is \(d=19-24=-5\). Using \(a_n=a+(n-1)d\), we get \(24+(n-1)(-5)=-11\). Thus, \(29-5n=-11\), which gives \(n=8\). Therefore, \((-11)\) is the eighth term. The seventh term is \((-6)\), so it is a close but incorrect option. Exam tip: In a decreasing AP, make sure to use a negative common difference.
The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Thus, \(a_5=12+(5-1)\times9=48\) and \(a_8=12+(8-1)\times9=75\). Therefore, \(a_5+a_8=48+75=123\). A value such as 117 may result from counting the common difference incorrectly. Exam tip: write \(n-1\) first while finding each term, then add the terms.
In (S_n=\frac{n}{2}[14+6(n-1)]), (n=11) gives (407) and (n=12) gives (480), so no option gives (539). In exams, check by substituting options.
Direct answer: option A, 6n − 3. In an arithmetic progression, consecutive terms differ by the constant d. From the fourth to the tenth term there are 10 − 4 = 6 equal steps, and the total increase is 57 − 21 = 36. Thus 6d = 36, so d = 6. To find the first term, move backward from a₄ by three steps: a₁ = 21 − 3(6) = 3. Now use aₙ = a₁ + (n − 1)d: aₙ = 3 + (n − 1)6 = 3 + 6n − 6 = 6n − 3. Check: a₄ = 24 − 3 = 21 and a₁₀ = 60 − 3 = 57. Option B gives 27 at n = 4, so its constant is wrong. Option C has common difference 5, not 6. Option D has difference 7 and also fails the given terms. Common confusion: the number of gaps between a₄ and a₁₀ is six, while the number of listed term positions is seven.
QUIZ COMPLETE
