Question

the three consecutive terms in AP whose sum is 54 and product is 5382

Answer 1

a-d+a+a+d=54

3a=54

a=18

second term is 18

a(a-d)(a+d)=5382

18(a^2-d^2)=5382

18^2-d^2=299

324-d^2=299

d^2=25

d=√25

d=5

so the three terms are 13,18,23

Answer 2

Answer:

Step-by-step explanation:

let the 3 terms Be a-d, a, a+d

A+a-d+a+d = 54

3a = 54 , a= 18

A(a+d)(a-d) = 5382

18(a² - D²) = 5382

324-d² = 5382/18

324-d² = 299

D² = 324-299

D² = 25

D = ±5

Ap formed when d=5

18,23,28......

Ap formed when d = -5

18, 13 , 7.....

Mark as brainliest