The (9)th term of an arithmetic progression is (46) and the (21)th term is (106). What is the sum of the first (21) terms?
Answer and explanation
Correct answer: 1176
For an AP, \(T_n=a+(n-1)d\). Thus, \(a+8d=46\) and \(a+20d=106\). Subtracting gives \(12d=60\), so \(d=5\) and \(a=6\). Therefore, \(S_{21}=\frac{21}{2}[2a+20d]=\frac{21}{2}[12+100]=1176\). Option 1113 can result from using \(21\times53\), but the average of the first and 21st terms is \(56\), not 53. Exam tip: when two terms are given, form equations first to find \(a\) and \(d\).
Frequently asked questions
What is the correct answer to this question?
1176
Why is this the correct answer?
For an AP, \(T_n=a+(n-1)d\). Thus, \(a+8d=46\) and \(a+20d=106\). Subtracting gives \(12d=60\), so \(d=5\) and \(a=6\). Therefore, \(S_{21}=\frac{21}{2}[2a+20d]=\frac{21}{2}[12+100]=1176\). Option 1113 can result from using \(21\times53\), but the average of the first and 21st terms is \(56\), not 53. Exam tip: when two terms are given, form equations first to find \(a\) and \(d\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.
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