Question

Four numbers are in A.P. The sum of first and fourth terms is 14 and the product of second and third is 45 find the numbers.​

Answer 1

Let ,

The four consecutive terms of AP be (a - 3d) , (a - d) , (a + d) , (a + 3d)

According to the question ,

The sum of first and fourth terms is 14

Thus ,

a - 3d + a + 3d = 14

2a = 14

a = 14/2

a = 7

And the product of second and third term is 45

Thus ,

(a - d)(a + d) = 45

a² + ad - ad - d² = 45

a² - d² = 45

(7)² - d² = 45

49 - d² = 45

d² = 4

d = ± 2

$\therefore$ The four consecutive terms of AP are 1 , 5 , 9 , 13 or 13 , 9 , 6 , 1