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

Mathematics
JAMB 1981

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
  • A. 12
  • B. 27
  • C. 9
  • D. 4
  • E. 36
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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

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