Browse Questions

# Find the sum of all the numbers between $200$ and $400$ which are divisible by $7$.

$\begin{array}{1 1}9030 \\ 8127 \\ 8428 \\ 8729 \end{array}$

Toolbox:
• $n^{th}$ term of an A.P.$=t_n=a+(n-1)d$
• sum of $n$ terms of an A.P.$=\large\frac{n}{2}$$(l+a) where l=t_n Numbers between 200 and 400 divisible by 7 forms the sequence 203,210,217........399 This sequence is an A. P. with first term a=203 and common difference d=7 The last term or n^{th} term of this sequence is t_n=399 We know that n^{th} term of an A.P. = a+(n-1)d \Rightarrow\:a+(n-1)d=399 \Rightarrow\:203+(n-1)7=399 \Rightarrow\:(n-1)7=399-203=196 \Rightarrow\:n-1=\large\frac{196}{7}$$=28$
or $n=29$
$\Rightarrow\:$The sum of the required numbers =
$203+210+217+......399=\large\frac{n}{2}$$(t_n+a) =\large\frac{29}{2}$$(399+203)=29\times301=8729$