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11. Find five numbers in A.P., such that their sum is 20 and the product of the first and the last is 15. ​

User Leandrojmp
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2 Answers

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Call the 5 numbers x, x + y, x + 2y, x + 3y, x + 4y, where x is the value of the first number and y is the constant value difference.

Since their sum is 20, 5x + 10y = 20.

Since the product of the first and last number is 15, x^2 + 4xy = 15.

From 5x + 10y = 20, we can see that x + 2y = 4.

-> (x+2y)^2 = 16 -> x^2 + 4xy + 4y^2 = 16 -> 4y^2 = 1.

-> y = 1/2 or -1/2

Replace back into the original equation : 5x + 10.(1/2) = 20 or 5x + 10.(-1/2) = 20

-> 5x + 5 = 20 / 5x + (-5) = 20

-> x = 3 / x = 5.

From the 4 combinations of x and y, we see that (x;y) = (5; -1/2) and (3; 1/;2) would satisfy the second condition.

So, the five numbers that we need to find has 2 different combinations :

(5; 9/2; 4; 7/2; 3) or vice versa (sum = 20)

User Tstenner
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5 votes

Answer: See below

Explanation:

I'm assuming that A.P is a list...

The prime factorization of 15 is: 1, 3, 5, 15, which means that the first and last can only be 1 and 15, or 3 and 5.

However, If it were 1 and 15, this would mean that the sum of the other 3 numbers equals 4. Not sure if repetition is allowed.

Option 1: 1 and 15

Other numbers are 1, 1, and 2

Option 2: 3 and 5

Other numbers add up to 13

Only one number between 3 and 5 though, which is 4. Do you know if the list is numbered from least to greatest?

User Ash Wilson
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