A newspaper seller sells (90) newspapers on the first day and (10) more newspapers each next day. On which day will (230) newspapers be sold?
Answer and explanation
Correct answer: 15
Here, the first term is 90 and the common difference is 10. Therefore, the number of newspapers sold on the nth day is
\(a_n=90+(n-1)\times10\). Putting
\(90+(n-1)\times10=230\) gives
\(n-1=14\), so
\(n=15\). Day 14 is a close distractor, but only 220 newspapers would be sold that day. Exam tip: equate the target value to the nth-term formula of the AP and solve for n.
Frequently asked questions
What is the correct answer to this question?
15
Why is this the correct answer?
Here, the first term is 90 and the common difference is 10. Therefore, the number of newspapers sold on the nth day is
\(a_n=90+(n-1)\times10\). Putting
\(90+(n-1)\times10=230\) gives
\(n-1=14\), so
\(n=15\). Day 14 is a close distractor, but only 220 newspapers would be sold that day. Exam tip: equate the target value to the nth-term formula of the AP and solve for n.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Word problems based on APs.
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