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

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