Question

The first and fifth terms of an A.P. are -14 and 2 respectively and the sum of its $n$ terms is 40. The value of $n$ is

• A. 8
• B. 12
• C. 10
• D. 13

Hint:

Always use the sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \) to connect multiple knowns in an arithmetic progression.

Answer 1

The Correct Option is C

We are given:
First term \( a = -14 \)
Fifth term \( a + 4d = 2 \)
Sum of $n$ terms: \( S_n = 40 \)
Step 1: Find the common difference \( d \)
Using the formula for the $n^{\text{th}}$ term: \[ a + 4d = 2
-14 + 4d = 2
4d = 16 \Rightarrow d = 4 \] Step 2: Use the formula for the sum of an A.P.
\[ S_n = \frac{n}{2}[2a + (n-1)d] \] Substitute values: \[ 40 = \frac{n}{2}[2(-14) + (n-1)(4)]
40 = \frac{n}{2}[-28 + 4n - 4]
40 = \frac{n}{2}[4n - 32]
80 = n(4n - 32)
4n^2 - 32n - 80 = 0 \Rightarrow n^2 - 8n - 20 = 0 \] Step 3: Solve the quadratic \[ n^2 - 8n - 20 = 0
n = \frac{8 \pm \sqrt{(-8)^2 + 4(1)(20)}}{2} = \frac{8 \pm \sqrt{64 + 80}}{2} = \frac{8 \pm \sqrt{144}}{2} = \frac{8 \pm 12}{2} \] \[ n = 10 \text{ or } n = -2 \quad (\text{Reject negative value}) \] So, \( n = 10 \)