Question

1.3

Find the three primes whose sum and product are 15 and 105 respectively.​

Answer 1

Answer:   here is your answer,

(a – d ) + a + (a + d ) = 15

=> 3a = 15

=>  a = 5

Again,

The three numbers in ap whose product is 105.

So,

(a – d) a (a + d) = 105

a(a^2 – d^2) = 105

5(25 – d^2) = 105

25 – d^2 = 21

d^2 = 4

d = ± 2

When , d = 2

The three terms of the A.P

= (5 – 2), 5 and (5 + 2)

= 3, 5 and 7

And when, d = – 2

The three terms of the A.P = (5 + 2) 5, and (5 – 2)

= 7, 5 and 3.