Question

Find the three consecutive terms of an
A.P. If their sum is -3 and product of
their Square is 64​

Answer 1

Answer:

let the terms of ap be : a-d , a , a+d

Sum of terms = a+d +a + a-d

-3 = 3a+d-d

-3 = 3a

a = -1.

Product of their squares = (a+d)².(a)².(a-d)²

• 64 = 8² = [ (a+d).(a).(a-d) ]²
• 8 = (a).(a+d).(a-d)
• 8 = a.(a²- d²)
• 8 = (-1) ( (-1)² - d² )
• -8 = 1 - d²
• d² = 9

So d = +/- 3.

The terms can be : a=-1, d=3 : [ 2 , -1 , -3 ]

Or, a= -1, d = -3 : [ -4 , -1 , 2 ]

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