Question

Find the three numbers in AP whose sum
30 and their product
is
is 990.​

Answer 1

Answer:

the greatest number in AP is 13.

Step-by-step explanation:

Given : Three numbers are in Arithmetic Progression (AP) whose sum is 30 and the product is 910.

To find : The greatest number in the AP is?

Solution :

Let the Arithmetic Progression (AP) is (a-d),a,(a+d)

AP sum is 30.

i.e. a-d+a+a+d=30a−d+a+a+d=30

3a=303a=30

a=10a=10

AP product is 910.

i.e. (a-d)\times a\times (a+d)=910(a−d)×a×(a+d)=910

(a^2-d^2)a=910(a

2

−d

2

)a=910

Substitute a=10,

(10^2-d^2)10=910(10

2

−d

2

)10=910

100-d^2=91100−d

2

=91

d^2=100-91d

2

=100−91

d^2=9d

2

=9

d=\pm3d=±3

Now, If a=10 and d=3,

AP is (10-3),10,(10+3)

AP is 7,10,13

Greatest is 13.

Now, If a=10 and d=-3,

AP is (10+3),10,(10-3)

AP is 13,10,7

Greatest is 13.

Therefore, The greatest number in the AP is 13.