The average of 5 consecutive integers starting with ‘m’ is n. What is the average of 6 consecutive integers starting with (m + 2)?
A. $\Large\frac{(2 n+5)}{2}$ B. (2n + 2) C. (n + 3) D. $\Large\frac{(2 n+9)}{2}$ Answer: Option AShow Answer
Solution(By Apex Team)
According to the question,
Let M = 1
∴ 5 consecutive integers are = 1, 2, 3, 4, 5
$\begin{array}{l}\therefore\frac{1+2+3+4+5}{5}=n\\
n=\frac{15}{5}\\
=3\end{array}$
∴ 6 consecutive integers starting with (m + 2) are = 3, 4, 5, 6, 7, 8
$\begin{array}{l}\therefore\Large\frac{3+4+5+6+7+8}{6}\\
=\Large\frac{33}{6}\\
=\Large\frac{11}{2}\end{array}$
Now check from option to put n = 3
$\begin{array}{l}\text{Option : (A) }\\
\Large\frac{(2n+5)}{2}\\
=\Large\frac{2\times3+5}{2}\\
=\frac{11}{2}\ \left(\text{satisfied}\right)\end{array}$
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