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3 years ago

The sum of five consective even numbers is less than 100.Find the maximum sum of these five numbers

Mathematics

2 answers:

Snowcat [4.5K]3 years ago

7 0

X+x+2+x+4+x+6+x+8 < 100
5x + 20 < 100
5x < 80
x < 16

so the 5 numbers are 14,16,18,20,22 and maximum sum = 90

Ludmilka [50]3 years ago

4 0

100/5 = 20

would be the middle number so 18 +19+20+21+22 = 100

since you want it to be less than 100 use 17 instead of 22

so you have 17 +18+19+20+21 = 95

95 is the maximum sum

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3 years ago

(a) Find the common difference of an A.P. whose first term is 5 and the number of terms is 30 and sum 1455.

Alexxandr [17]

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3 years ago

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2 years ago

Explain why any of the four operations placed between two terms 5 and -3 sqrt (8) will result in an irrational number

Gnoma [55]

Sum/difference:

Let

$x = 5 + (-3\sqrt{8}) = 5-3\sqrt{8}$

This means that

$3\sqrt{8} = 5-x \iff \sqrt{8} = \dfrac{5-x}{3}$

Now, assume that $x$ is rational. The sum/difference of two rational numbers is still rational (so 5-x is rational), and the division by 3 doesn't change this. So, you have that the square root of 8 equals a rational number, which is false. The mistake must have been supposing that $x$ was rational, which proves that the sum/difference of the two given terms was irrational

Multiplication/division:

The logic is actually the same: if we multiply the two terms we get

$x = -15\sqrt{8}$

if again we assume x to be rational, we have

$\sqrt{8} = -\dfrac{x}{15}$

But if x is rational, so is -x/15, and again we come to a contradiction: we have the square root of 8 on one side, which is irrational, and -x/15 on the other, which is rational. So, again, x must have been irrational. You can prove the same claim for the division in a totally similar fashion.

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2 years ago