Question

find three consecutive terms in A.P whodmse sum is -3 and the product of their cubes is 512(take the value of d positive)​

Answer 1

Answer:

the 3 no.s are -4, -1 , 2

Step-by-step explanation:

let (a-d), a, and (a+d) be the 3 consecutive terms of the AP

=> given, (a-d)+a+(a+d) = -3

ie., 3a = -3

=> a = -3/3 = -1

also, (a-d)³*a³*(a+d)³ = 512

=>  (-1-d)³*(-1)³*(-1+d)³ = 512

=> (-1)3* (d+1)³*(-1)³*(d-1)³ = 512

=> (d+1)³*(d-1)³ = 512

=> (d²-1)(d²-1)(d²-1) = 512

=> (d²-1)³ = 512

=> (d²-1)³ = 8³

=> d²-1 = 8

=> d² = 8+1

=> d² = 9

=> d = ±3

so, take a=-1 with d=3  (given to take d as a positive integer)

=> the 3 no.s are (-1-3), -1 , (-1+3)

=> the 3 no.s are -4, -1 , 2