If (a_1=12) and it is not (a_n=3a_{n-1}+2) but an AP with (a_n=12+(n-1)d), and (a_9=68), what is (a_{22})?
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Answer and explanation
Correct answer: 159
Using the AP formula, \(a_9=12+(9-1)d\). Thus, \(68=12+8d\), so \(d=7\). Now \(a_{22}=12+(22-1)\times7=12+147=159\). Choosing 166 would result from using an incorrect common difference instead of 7. Exam tip: first find \(d\) from the given term, then use \(n-1\) for the required term.
Frequently asked questions
What is the correct answer to this question?
159
Why is this the correct answer?
Using the AP formula, \(a_9=12+(9-1)d\). Thus, \(68=12+8d\), so \(d=7\). Now \(a_{22}=12+(22-1)\times7=12+147=159\). Choosing 166 would result from using an incorrect common difference instead of 7. Exam tip: first find \(d\) from the given term, then use \(n-1\) for the required 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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