Question

find the three term in ap such that their sum is 3 and product is -8

Answer 1

Heya !!




Let Required numbers are ( a - d ) , a , ( a + d ).






Then,


a - d + a + a + d = 3



3a = 3



a = 1



And,



(a - d ) a ( a + d ) = -8



a ( a² - d² ) = -8



1 ( 1 - d² ) = -8




-d² = -8-1




-d² = -9



d = √9 = 3






First term ( a ) = 1


Common difference (d ) = 3.





Hence,


Required numbers are ( -2 , 1 , 4 ).

Answer 2

Let the three terms in AP be a-d, a, a+d

Where a is first term of AP and d is common difference between the terms

Now, Given that sum of terms is 3

Sum of terms = 3

( a-d) + a + (a+d) = 3)

3a = 3

a = 1

Again,

Product of terms = (-8)

(a-d)*a(a+d) = -8

(1-d)*1 *(1+d) = -8

(1² -d² ) = -8

-d² = -8 - 1

-d² = -9

d²= 9

d = +3 or -3

If d = +3 and a=1

Then terms will be -2 , 1 , 4

if d = -3 and a = 1

Then terms will be 4 , 1 , -2