Question

the sum of 3 numbers in a.p is 15 and the product of the first and last is 21 find the numbers

Answer 1

Hey there !

Thanks for the question !

Here's the answer !

Let the three numbers be: ( a - d ), ( a ), ( a + d )

Given that the sum of the terms is 15 and the product of first and last term is 21.

So let us solve it step by step.

Sum = a - d + a + a + d = 15

=> 3a + d - d = 15

=> 3a = 15

=> a = 15 / 3 = 5

So the central term is 5.

Product of last and first term is : ( a + d ) ( a - d )

This is of the form ( a - b ) ( a + b ) = a² - b²

=> ( a + d ) ( a - d ) = a² - d²

=> a² - d² = 21

We know that a = 5. Substituting that we get,

=> 5² - d² = 21

=> 25 - d² = 21

=> 25 - 21 = d²

=> 4 = d²

=> d = √ 4 => +2 or -2.

So if d = +2, we get the terms to be:

5 - 2 , 5 , 5 + 2 = 3, 5, 7

If d = -2, then we get the terms to be:

5 - ( -2 ), 5 , 5 + ( -2 ) = 7, 5, 3

Hence in both the cases the numbers are same.

Hence the numbers are 3, 5 and 7.

Hope my answer helped !

Answer 2

Let the three numbers in an AP be a - d,a,a + d.

Given that sum of three numbers in an AP is 15.

= > a - d + a + a + d = 15

= > 3a = 15

= > a = 5

Given that product of first and last term is 21.

= > (a - d) * (a + d) = 21

= > a^2 - d^2 = 21

= > 5^2 - d^2 = 21

= > -d^2 = 21 - 25

= > -d^2 = -4

= > d = +2,-2.

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When a = 5, d = 2:

= > a - d = 3

= > a = 5

= > a + d = 7.

When a = 5, d = -2

= > a - d = 5 + 2 = 7

= > a = 5

= > a + d = 5 - 2 = 3.

Therefore, the numbers are 3,5,7 (or) 7,5,3.

Hope this helps!