Question 1090952
 When a group of 50 consecutive multiples of 7 is added 
the answer is 12075. what's the smallest and the biggest number
<pre>
Let k the smallest integer such that 7k is the smallest multiple
of 7 such that the sum of the series

7k+(7k+7)+(7k+14)+ &#8729;&#8729;&#8729; (to 50 terms) = 12075

Th formula for the sum of an arithmetic series is

{{{S[n]=expr(n/2)(2a[1]+(n-1)^""*d)}}}, where n=50, a<sub>1</sub>=7k, d=7

{{{S[50]=expr(50/2)(2(7k)+(50-1)^""*7)}}}

{{{S[50]=(25)(14k+49*7)}}}

{{{S[50]=(25)(14k+343)}}}

And since S<sub>50</sub>=12075

{{{(25)(14k+343)=12075}}}

Divide both sides by 25

{{{14k+343=483}}}

{{{14k=140}}}

{{{k=10}}} 

The smallest integer = 7k = 7(10) = 70
The largest integer is the 50th term

{{{a[n]=a[1]+(n-1)*d}}}

{{{a[50]=70+(50-1)*7}}}

{{{a[50]=70+49*7}}}

{{{a[50]=70+343}}}

{{{a[50]=413}}}

Answers: smallest = 70, biggest = 413

Edwin</pre>