Find the average of ten consecutive multiples of 12, starting from the fifth multiple.
Correct Answer: Option C
We know that,
Average = \(\frac{{{\bf{sum}}\ {\bf{of}}\ {\bf{multiples}}}}{{{\bf{no}}.\ {\bf{of}}\ {\bf{multiples}}}}\)
\(= \frac{{\left( {12 \times 5} \right) + \left( {12 \times 6} \right) + \ldots \ldots \ldots . + \left( {12 \times 14} \right)}}{{10}}\)
=12/10 (5+6+ ………+14)
=12/10 (1+2+3+ ………+14-(1+2+3+4))
\(= \frac{{12}}{{10}}\left( {\frac{{14 \times 15}}{2} - \frac{{4 \times 5}}{2}} \right)\)
∵ Sum of first n numbers = \(\frac{{n\left( {n + 1} \right)}}{2}\)
∴ Required average \(= \frac{{12}}{{10}} \times \left( {105 - 10} \right) = 114\)
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