Question

The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
A. 5
B. 10
C. 12
D. 14

Answer 1

I think answer is option A. 5 is the correct answer if not sorry

Answer 2

D. 14

Step-by-step explanation:

Given: The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406.

Solution:

n = ?

Sum = 406

a = 3

d = 7-3 = 4

So we have Sum = n/2 (2a +(n-1)d)

406 = n/2 (6 +(n-1)4)

406 *2 = n(6 +4n -4)

406 *2 = 6n + 4n² -4n

4n² + 2n -(406*2) = 0

2n² + n - 406 = 0

2n² - 28n +29n - 406 = 0

2n(n - 14) +29 (n-14) = 0

 (2n +29)(n-14) = 0

n can be -29/2 or +14

n can't be negative, so n = 14

Option D is the answer.