Find the sum of the first 20 terms of the series 8, 12, 16, ....., 96
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Correct Answer: Option D
Explanation:
8, 12, 16, .....96
a = 8, d = 4, l = 96, n = 20
\(S_{20} = \frac{n}{2}(2a + (n-1)d\)
\(S_{20} = \frac{20}{2}((2\times 8) + (20-1)\times 4)\)
\(S_{20} = 10(16 + (19\times 4)) = 10 \times 92\)
=920
8, 12, 16, .....96
a = 8, d = 4, l = 96, n = 20
\(S_{20} = \frac{n}{2}(2a + (n-1)d\)
\(S_{20} = \frac{20}{2}((2\times 8) + (20-1)\times 4)\)
\(S_{20} = 10(16 + (19\times 4)) = 10 \times 92\)
=920