SOLUTION: When a group of 50 consecutive multiples of 7 is added the answer is 12075. what's the smallest and the biggest number

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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
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20066) About Me  (Show Source):
You can put this solution on YOUR website!
When a group of 50 consecutive multiples of 7 is added
the answer is 12075. what's the smallest and the biggest number
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)+ ∙∙∙ (to 50 terms) = 12075

Th formula for the sum of an arithmetic series is

S%5Bn%5D=expr%28n%2F2%29%282a%5B1%5D%2B%28n-1%29%5E%22%22%2Ad%29, where n=50, a1=7k, d=7

S%5B50%5D=expr%2850%2F2%29%282%287k%29%2B%2850-1%29%5E%22%22%2A7%29

S%5B50%5D=%2825%29%2814k%2B49%2A7%29

S%5B50%5D=%2825%29%2814k%2B343%29

And since S50=12075

%2825%29%2814k%2B343%29=12075

Divide both sides by 25

14k%2B343=483

14k=140

k=10 

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

a%5Bn%5D=a%5B1%5D%2B%28n-1%29%2Ad

a%5B50%5D=70%2B%2850-1%29%2A7

a%5B50%5D=70%2B49%2A7

a%5B50%5D=70%2B343

a%5B50%5D=413

Answers: smallest = 70, biggest = 413

Edwin

Answer by ikleyn(52921) About Me  (Show Source):
You can put this solution on YOUR website!
.
Use the formula for the sum of the first 50 terms of an arithmetic progression with the common difference 7,

%28a%5B1%5D%2Ba%5B50%5D%29%2F2.50 = 12075.     (1)


Since a%5B50%5D = a%5B1%5D%2B7%2A49, you can rewrite (1) in the form


%28a%5B1%5D+%2B+a%5B1%5D%2B7%2A49%29%2F2.50 = 12075,   or

%282a%5B1%5D+%2B+7%2A49%29.50 = 24150,

2a%5B1%5D = 24150%2F50+-+7%2A49  ====>  a%5B1%5D = %28483+-+7%2A49%29%2F2 = 70.


Thus you found the smallest term of the AP. It is 70.


Then the biggest term is 70 + 7*49 = 413.


Happily, the first term is multiple of 7 and the common difference is 7, so all the terms of the progression 
are consecutive integers multiple of 7.

Solved.


There is a bunch of lessons on arithmetic progressions in this site:
    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
    - Mathematical induction and arithmetic progressions
    - One characteristic property of arithmetic progressions
    - Solved problems on arithmetic progressions


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".