In the AP (156,142,128,\ldots), which is the last term greater than (-120)?
Answer and explanation
Correct answer: (-110)
Here, the first term is \(a=156\) and the common difference is \(d=-14\). The \(n\)th term is \(a_n=156+(n-1)(-14)=170-14n\). For a term greater than \((-120)\), \(170-14n>-120\), so \(n<20.71\). Thus the greatest integral value is \(n=20\), giving \(a_{20}=-110\). The next term, \(a_{21}=-124\), is not greater than \((-120)\). Exam tip: In a decreasing AP, always check the next term after finding the candidate last term.
Frequently asked questions
What is the correct answer to this question?
(-110)
Why is this the correct answer?
Here, the first term is \(a=156\) and the common difference is \(d=-14\). The \(n\)th term is \(a_n=156+(n-1)(-14)=170-14n\). For a term greater than \((-120)\), \(170-14n>-120\), so \(n<20.71\). Thus the greatest integral value is \(n=20\), giving \(a_{20}=-110\). The next term, \(a_{21}=-124\), is not greater than \((-120)\). Exam tip: In a decreasing AP, always check the next term after finding the candidate last term.
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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