Q. The arithmetic mean of 7 consecutive integers starting with $a$ is $m$. Then, the arithmetic mean of 11 consecutive integers starting with $a + 2$ is
A) $2a$
B) $2m$
C) $a + 4$
D) $m + 4$
Explanation
(d)
$\frac{a + (a + 1) + (a + 2) + ... + (a + 6)}{7} = m$
$\Rightarrow 7a + 21 = 7m$
$\Rightarrow a + 3 = m$
$\therefore\ \ \ \ \frac{(a + 2) + (a + 3) + (a + 4) + \ldots + (a + 12)}{11}$
$= \frac{11a + 77}{11}$
$= a + 7$
$= m - 3 + 7$
$= m + 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.