For the given series 10, 7, 4, ..., −62, find the 11th term from the end.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −32
The series is an A.P. with common difference \(10-7=-3\). Counting from the end is counting from the last term in the reverse direction, so the \(n\)th term from the end is \(l-(n-1)d\), where \(l\) is the last term.
Last term \(l=-62\), \(d=-3\), \(n=11\)
11th term from the end \(=-62-10\times(-3)=-62+30=-32\)