If (2a+3, 5a-1, 11a-13) are in an arithmetic progression, what is the second term?
Answer and explanation
Correct answer: \(\frac{37}{3}\)
In an arithmetic progression, twice the middle term equals the sum of the first and third terms. Thus, \(2(5a-1)=(2a+3)+(11a-13)\). This gives \(10a-2=13a-10\), so \(a=\frac{8}{3}\). Hence, the second term is \(5a-1=5\times\frac{8}{3}-1=\frac{37}{3}\). The value \(14\) results from the incorrect calculation \(a=3\). Exam tip: for three AP terms, use \(2b=a+c\) directly.
Frequently asked questions
What is the correct answer to this question?
\(\frac{37}{3}\)
Why is this the correct answer?
In an arithmetic progression, twice the middle term equals the sum of the first and third terms. Thus, \(2(5a-1)=(2a+3)+(11a-13)\). This gives \(10a-2=13a-10\), so \(a=\frac{8}{3}\). Hence, the second term is \(5a-1=5\times\frac{8}{3}-1=\frac{37}{3}\). The value \(14\) results from the incorrect calculation \(a=3\). Exam tip: for three AP terms, use \(2b=a+c\) directly.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Introduction to APs and common difference..
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