Question

Find three consecutive terms in an AP whose sum is 27 and product is 540​

Answer 1

Step-by-step explanation:

Let three nos be

(a-d),a,(a+d)

where (a-d) is smallest and (a+d) is biggest

according to condition.

a+a+d+a-d = 27......(1)

3a = 27

a = 9

a× (a+d)×(a-d) = 540

a× a^2 - d^2 = 540

a^3 - d^2 = 540

d^2 = 729 - 540

d ~ 13

Answer 2

let the terms be a-d,a,a+d

Using 1st condition,

a-d+a+a+d=27

3a=27

a=9

Using the 2nd condition,

(a-d)(a)(a+d)=540

Substituting value of a,

(9-d)(9+d)(9)=504

(81-d^2)(9)=504

81-d^2=540/9=56

-d^2=-25

Hence, d=5

So the numbers are 9-5,9,9+5=4,9,14

You might have noticed the RHS in the second condition is 504 not 540 because you typed it wrong (not being rude or anything). Please check the question again.