In an AP, (a_1+a_4=38) and (a_7=44). What is (a_{16})?
Answer and explanation
Correct answer: 94
Let the first term be \(a\) and the common difference be \(d\). Then \(a_1+a_4=a+(a+3d)=2a+3d=38\), while \(a_7=a+6d=44\). Multiplying the second equation by 2 gives \(2a+12d=88\). Subtracting the first equation gives \(9d=50\), so \(d=\frac{50}{9}\). Hence, \(a_{16}=a_7+9d=44+9\times\frac{50}{9}=94\). The value 80 results from incorrectly taking \(d=4\). Exam tip: eliminate the common term by multiplying one AP equation suitably before subtracting.
Frequently asked questions
What is the correct answer to this question?
94
Why is this the correct answer?
Let the first term be \(a\) and the common difference be \(d\). Then \(a_1+a_4=a+(a+3d)=2a+3d=38\), while \(a_7=a+6d=44\). Multiplying the second equation by 2 gives \(2a+12d=88\). Subtracting the first equation gives \(9d=50\), so \(d=\frac{50}{9}\). Hence, \(a_{16}=a_7+9d=44+9\times\frac{50}{9}=94\). The value 80 results from incorrectly taking \(d=4\). Exam tip: eliminate the common term by multiplying one AP equation suitably before subtracting.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the $n$th term of an AP.
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