The first term of an Arithmetic progression is 3 and the fifth term is 9….

The first term of an Arithmetic progression is 3 and the fifth term is 9. Find the number of terms in the progression if the sum is 81

Correct Answer: Option C

Explanation

1st term a = 3, 5th term = 9, sum of n = 81
nth term = a + (n – 1)d, 5th term a + (5 – 1)d = 9
3 + 4d = 9
4d = 9 – 3
d = (frac{6}{4})
= (frac{3}{2})
= 6
S_n = (frac{n}{2})(6 + (frac{3}{4})n – (frac{3}{2}))
81 = (frac{12n + 3n^2}{4}) – 3n
= (frac{3n^2 + 9n}{4})
3n² + 9n = 324
3n² + 9n – 324 = 0
By almighty formula positive no. n = 9
= 3