# Question #09e08

Feb 9, 2017

Let n be the smallest of 6 consecutive number.So sum of 6 consecutive numbers forming an AP series of common difference 1 will be

$= \frac{6}{2} \left(2 \cdot n + \left(6 - 1\right) \cdot 1\right) = 6 n + 15$

By the given condition

$6 n + 15 = 837$

$\implies n = \frac{837 - 15}{6} = 137$

So 6th number$= 137 + \left(6 - 1\right) \cdot 1 = 142$