The sum of the first four terms of an A.P. is 56 and the sum of its last four terms is 112. If its first term is 11, then find the number of terms.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 11
In an A.P. with common difference \(d\), the first four terms are \(a, a+d, a+2d, a+3d\) and the last four are \(l-3d, l-2d, l-d, l\), where \(l\) is the last term. The two sums give \(d\) and then \(l\).
The common difference (sum of the first four terms):
The four terms add up to \(4a+6d=56\)
With \(a=11\): \(44+6d=56\), so \(d=2\)
The last term (sum of the last four terms):
The last four terms add up to \(4l-6d=112\)
With \(d=2\): \(4l-12=112\), so \(l=31\)
The number of terms (\(l=a+(n-1)d\)):
Last term 31, first term 11, difference 2: \(31=11+(n-1)\times2\)
\(n-1=10\), so the A.P. has 11 terms