Question

Find the three terms of an AP such that their sum and product are -3 and 15 respectively.

Answer 1

Solution :- $\sf$

• Let the 3 term in A.P be = ( a-d ) , a , ( a+d )

Given :-

• Sum of 3 term = -3
• Product is = 15

According to question :-

• sum of 3 term in A.P = -3

So,

→ ( a-d ) + a + ( a+d ) = -3

→ 3a = -3

→ a = -3/3

→ a = -1

• Product of 3 term in an A.P = 15

So,

→ ( a-d ) , a , ( a+d ) = 15 ( putting value of a )

→ (-1 - d ) × -1 ( -1 + d ) = -15

→ (-1 - d ) ( -1 + d ) = -15

→ (-1 )² - ( d )² = -15

→ 1 - 15 = d²

→ 16 = d²

→ √16 = d

→ 4 = d

Hence,

• a - d = -1 - 4 = -5
• a = -1
• a + d = -1 + 4 = 3

The number are = -5, -1 , 3

Answer 2

Step-by-step explanation:

-5,-1,3

hope it helps u