Question

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.

• A. 7
• B. 11
• C. 13
• D. 17

Answer 1

The Correct Option is B

To solve the problem, let's analyze the arithmetic progression (A.P.) given and use the properties of A.P.

Given:

• Sum of the first 4 terms = 56
• Sum of the last 4 terms = 112
• First term \(a = 11\)

Let \(d\) be the common difference and \(n\) be the number of terms. The general nth term of an A.P. is given by:

\(a_n = a + (n-1)d\)

First 4 terms of the A.P.: \(a, a+d, a+2d, a+3d\)
Their sum is:

\(a + (a+d) + (a+2d) + (a+3d) = 4a + 6d = 56\)
Substituting \(a = 11\):

\(4(11) + 6d = 56\)
\(44 + 6d = 56\)
\(6d = 12\)
\(d = 2\)

Last 4 terms of the A.P.: \(a_{n-3}, a_{n-2}, a_{n-1}, a_n\)
Their sum is:

\((a+(n-4)d) + (a+(n-3)d) + (a+(n-2)d) + (a+(n-1)d) = 4a + (4n-10)d = 112\)

Substitute \(a = 11\) and \(d = 2\):

\(4(11) + (4n-10)(2) = 112\)
\(44 + (8n-20) = 112\)
\(8n + 24 = 112\)
\(8n = 112 - 24\)
\(8n = 88\)
\(n = 11\)

Thus, the number of terms in the A.P. is 11.