SOLUTION: Find the four numbers in A.P. whose sum is 50 and in which the greatest number is 4 times the least.

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Question 1135721: Find the four numbers in A.P. whose sum is 50 and in which the greatest number is 4 times the least.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52807) About Me  (Show Source):
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From the statement  "the greatest number is 4 times the least" we get


    4%2Aa%5B1%5D = a%5B1%5D%2B3d  ====>


    3%2Aa%5B1%5D = 3d  ====>  a%5B1%5D = d.


From the statement  "the sum is 50"  we get then


    a%5B1%5D + %28a%5B1%5D%2Bd%29 + %28a%5B1%5D%2B2%2Ad%29 + %28a%5B1%5D%2B3d%29 = 50,   or


    4%2Aa%5B1%5D+%2B+6d = 50,   or   4d + 6d = 50,   or  10d = 50,  i.e.  d = 50/10 = 5.


Thus  a%5B1%5D = 5   and  d = 5.


The progression (the 4 terms) are  5, 10, 15 and 20.   ANSWER

Solved.


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


If the sum of 4 numbers in AP is 50, then the sum of the first and last is 25, and the sum of the 2nd and 3rd is 25.

We are told that the largest number is 4 times the smallest, so call them a and 4a.

Then we know a+4a = 5a = 25; so a=5.

Then the largest and smallest numbers are a=5 and 4a=20.

The difference of 15 between the largest and smallest of the 4 numbers is 3 times the common difference, so the common difference is 5. That makes the other two numbers 10 and 15.

So the 4 numbers are 5, 10, 15, and 20.

Note we don't know what the actual AP is; it could be either

5, 10, 15, 20

or

20,15,10,5