If \(a, b, c, d, e\) and \(f\) are six consecutive odd numbers, their average is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(a+5\)
Consecutive odd numbers go up in steps of 2, so each number can be written in terms of the first number \(a\).
The numbers are \(a,\ a+2,\ a+4,\ a+6,\ a+8,\ a+10\)
Sum \(=6a+(2+4+6+8+10)=6a+30\)
\(\displaystyle \text{Average}=\frac{6a+30}{6}=a+5\)
The average is \(a+5\). For example, 1, 3, 5, 7, 9, 11 have average 6, which is \(1+5\).
Why the other options are incorrect
Option (1) \(4(a+b)\) — Incorrect. For 1, 3, 5, 7, 9, 11 it gives \(4\times4=16\), not the average 6.
Option (2) \(\displaystyle \frac{abcdef}{5}\) — Incorrect. An average uses the sum, not the product; for 1, 3, 5, 7, 9, 11 it gives 2079.
Option (4) \(5(a+b+c+d+e+f)\) — Incorrect. The sum must be divided by 6, not multiplied by 5; for 1, 3, 5, 7, 9, 11 it gives 180.