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The sum of first five multiples of 3 is:
Answer & Solution
Correct Answer:
Option
A
$${S_n} = \left[ {2a + \left( {n - 1} \right)d} \right] \times \frac{n}{2}$$
$$ \Rightarrow {S_5} = \left[ {2 \times 3 + \left( {5 - 1} \right)3} \right]$$ $$ \times \frac{5}{2}$$
$$\eqalign{ & \Rightarrow {S_5} = \left[ {6 + 12} \right] \times \frac{5}{2} \cr & \Rightarrow {S_5} = 18 \times \frac{5}{2} \cr & \Rightarrow {S_5} = 9 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 45 \cr} $$
$$ \Rightarrow {S_5} = \left[ {2 \times 3 + \left( {5 - 1} \right)3} \right]$$ $$ \times \frac{5}{2}$$
$$\eqalign{ & \Rightarrow {S_5} = \left[ {6 + 12} \right] \times \frac{5}{2} \cr & \Rightarrow {S_5} = 18 \times \frac{5}{2} \cr & \Rightarrow {S_5} = 9 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 45 \cr} $$
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