Question

sum of 3 consecutive term of an ap is 9 and product is 24 find the no​

Answer 1

Answer:

First you’ll have to find the first three consecutive terms of AP and see if they add up to nine with each squares adding up to 35. It’s a trial and error process.

In order to arrive at the answer you will have to square all the numbers that is lower than 9 and ensure that the result of each number squared is not greater than 35.

0*0 = 0

1*1 = 1

2*2 = 4

3*3 = 9

4*4 = 16

5*5 = 25

1+9+25 = 35

WITH ELIMINATION IT IS CLEAR THAT THE SQUARE OF 1, 3, 5 EACH ADDED TOGETHER IS 35 SO THEREFORE:

Three consecutive terms of AP is AP (1,3,5)

Answer 2

Step-by-step explanation:

Let the terms be a-d, a and a+d

Now,

a-d+a+a+d=9

3a =9

a=3

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Now,

(a-d) (a) (a+d) =24

replacing a

(3-d) (3+d) 3 =24

Using identity

( 3^2-d^2) 3=24

(9-d^2)3 =24

(9-d^2) =24/3

(9-d^2) =8

-d^2=8-9

-d^2 =-1

d^2 =1

d= 1

d = +1, -1

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