Question

If the sum of a certain \(n\) number of terms of the A.P. \(25, 22, 19, \ldots\) is \(116\), then the last term is:

• A. \(4\)
• B. \(3\)
• C. \(2\)
• D. \(-4\)

Hint:

For A.P. problems:

First find (n) using the sum formula
Then use (an = a + (n-1)d) to get the last term

Answer 1

The Correct Option is A

Step 1: Identify the first term and common difference. Given A.P.: \(25, 22, 19, \ldots\) \[ a = 25,\quad d = 22 - 25 = -3 \] Step 2: Use the formula for the sum of first \(n\) terms of an A.P. \[ S_n = \frac{n}{2}\,[2a + (n-1)d] \] Given \(S_n = 116\): \[ 116 = \frac{n}{2}\,[2(25) + (n-1)(-3)] \] Step 3: Simplify the expression. \[ 116 = \frac{n}{2}\,[50 - 3n + 3] \] \[ 116 = \frac{n}{2}\,(53 - 3n) \] Step 4: Solve for \(n\). \[ 232 = n(53 - 3n) \] \[ 3n^2 - 53n + 232 = 0 \] \[ (3n - 29)(n - 8) = 0 \] \[ n = 8 \quad (\text{valid integer solution}) \] Step 5: Find the last term. The \(n\)th term of an A.P. is: \[ a_n = a + (n-1)d \] \[ a_8 = 25 + 7(-3) = 25 - 21 = 4 \] Step 6: Final conclusion. The last term of the A.P. is \(\boxed{4}\).