Jambmaths question:
Find the sum of the first 18 terms of the series 3, 6, 9, -, -, -, 36.
Option A:
433
Option B:
635
Option C:
513
Option D:
505
Jamb Maths Solution:
$\begin{align} & a=3,\text{ }d=6-3,\text{ }n=18 \\ & {{S}_{n}}=\tfrac{n}{2}[2a+(n-1)d] \\ & {{S}_{18}}=\tfrac{18}{2}\left[ 2\cdot 3+(18-1)3 \right]=9[6+51] \\ & {{s}_{18}}=9\times 57=513 \\\end{align}$
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