In an AP, (a_6+a_{18}=156) and (a_{10}+a_{22}=252). What is (a_{34})?
Answer and explanation
Correct answer: 342
For an AP, \(a_n=a+(n-1)d\). Thus, \(a_6+a_{18}=2a+22d=156\) and \(a_{10}+a_{22}=2a+30d=252\). Subtracting the first equation from the second gives \(8d=96\), so \(d=12\). Now \(2a+22(12)=156\) gives \(a=-54\). Hence, \(a_{34}=a+33d=-54+33(12)=342\). The value \(324\) does not satisfy the two given conditions. Exam tip: subtract two such sum equations to eliminate \(a\) and find \(d\) quickly.
Frequently asked questions
What is the correct answer to this question?
342
Why is this the correct answer?
For an AP, \(a_n=a+(n-1)d\). Thus, \(a_6+a_{18}=2a+22d=156\) and \(a_{10}+a_{22}=2a+30d=252\). Subtracting the first equation from the second gives \(8d=96\), so \(d=12\). Now \(2a+22(12)=156\) gives \(a=-54\). Hence, \(a_{34}=a+33d=-54+33(12)=342\). The value \(324\) does not satisfy the two given conditions. Exam tip: subtract two such sum equations to eliminate \(a\) and find \(d\) quickly.
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.
