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
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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
Sn = \(\frac{n}{2}\)(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
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
Sn = \(\frac{n}{2}\)(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