Question

if the sum of three numbers in ap is 12 and their product is 28. then numbers are​

Answer 1

Answer:

Given:-

• Sum of three numbers in AP is 12
• Let take numbers as a-d, a, a+d
• So, a-d+a+a+d = 12

• Product of numbers is 28

Find:-

• The numbers?

Solution:-

Sum of numbers is 12

$\to{ \sf{a - d + a + a + d = 12}}$

${ \to{ \sf{3a = 12}}}$

${ \to{ \sf{a = \frac{12}{3} = 4 }}}$

So, The value of a is 4

Product of numbers is 28

$\to{ \sf{a - d(a)(a + d) = 28}}$

$\to{ \sf{4 - d \times 4 \times 4 + d = 28}}$

$\to{ \sf{(4 - d)(4 + d) = \frac{28}{4} }}$

$\to{ \sf{ {4}^{2} - {d}^{2} = 7 }}$

${ \to{ \sf{16 - {d}^{2} = 7}}}$

$\to{ \sf{ { - d}^{2} = 7 - 16}}$

$\to \sf{ { - d}^{2} = - 9 }$

$\to \sf{ d = \sqrt{9} }$

$\to \sf{d = 3}$

So, The value of d is 3

First number = a - d = 4-3 = 1

Second number = a = 4

Third number = a + d = 4+3 = 7

Therefore, the numbers are 1 ,4 ,7