Question

The five numbers in A.P with the sum 20 and the product of the first and last 15 are​

Answer 1

Let the first number = a - 2d

Let the second number = a - d

Let the third number = a

Let the fourth number = a + d

Let the fifth number = a + 2d

According to question :

(a - 2d) + (a - d) + (a) + (a + d) + (a + 2d) = 20

a + a + a + a + a - d - 2d + d + 2d = 20

5a = 20

a = 20/5

a = 4

Now,

(a + 2d) (a - 2d) = 15

a² - 4d² = 15

(4)² - 4(d²) = 15

16 - 4d² = 15

4d² = 16 - 15

d² = 1/4

d = 1/2

d = 0.5

Hence :

The first number = a - 2d = 4 - 2(0.5) = 4 - 1 = 3

The second number = a - d = 4 - 0.5 = 3.5

The third number = a = 4

The fourth number = a + d = 4 + 0.5 = 4.5

The fifth number = a + 2d = 4 + 2(0.5) = 4 + 1 = 5

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Answer 2

Since the sum is 20

let the terms be (a-2d),( a-1d),a,(a+1d),(a+2d)

According to the question

a-2d+a-1d+a+a+1d+a+2d=20

5a=20

a=20/5

a=4

According to the question

first term×last term=15

(a-2d)(a+2d)=15

(a)^2-(2d)^2=15

since a =4

(4)^2-(2d)^2=15

16-4(d^2)=15

16-15=4(d^2)

1=4(d^2)

d^2=1/4

d=(1/4)^1/2

d=1/2

Hence the terms are

a-2d=4-2*1/2

a-2d=3

a-1d=4-1×1/2

a-1d=(8-1)/2 through LCM

a-1d=7/2

a-1d=3.5

a=4

a+1d=4+1×1/2

a+1d=(8+1)/2 through LCM

a+1d=9/2

a+1d=4.5

a+2d=4+2*1/2

a+2d=5