Question

Three numbers are in Arithmetic Progression (AP) whose sum is 30 and the product is 910. Then the greatest number in the AP is

Answer 1

I think this would help you

Answer 2

Answer:

The greatest number in the 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. $a-d+a+a+d=30$

$3a=30$

$a=10$

AP product is 910.

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

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

Substitute a=10,

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

$100-d^2=91$

$d^2=100-91$

$d^2=9$

$d=\pm3$

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.