SOLUTION: When a group of 50 consecutive multiples of 7 is added the answer is 12 075. i) What is the smallest of these numbers? (6) ii) What is the largest of these numbers?

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: When a group of 50 consecutive multiples of 7 is added the answer is 12 075. i) What is the smallest of these numbers? (6) ii) What is the largest of these numbers?      Log On


   

Question 1091174: When a group of 50 consecutive multiples of 7 is added the answer is 12 075.
i) What is the smallest of these numbers? (6)
ii) What is the largest of these numbers?
Answer by htmentor(1343) About Me  (Show Source):
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Let the first term, a = 7k, where k is the first multiple
Then the 50th term will be 7(k+49)
This is an arithmetic sequence, but the sum starts with n=k rather than at n=1
The sum of 50 terms of this sequence is (50/2)*(7k + 7(k+49)) = 12075
Solve for k:
14k + 343 = 12075/25 = 483
14k = 140
Thus k = 10
i) The first term is 7*10 = 70
ii) The 50th term is 7(10+49) = 413
Check: The general formula for this sequence is a_n = 70 + 7(n-1)
The sum of the first 50 terms is (50/2)(70+413) = 12075