Question

Sum of the three consecutive terms of an ap is 15 and their product 120 find those terms​

Answer 1

Answer:

The three terms are 6,5,4 Or 4,5,6

Step-by-step explanation:

Let the consecutive terms in AP be a-d, a and a+d So

By 1st condition a-d+a+a+d=15

3a = 15

a = 15/3

a = 5

By 2nd condition (a-d)×a×(a+d)= 120

a(a²-d²) = 120

5(5²-d²) = 120

5(25-d²) = 120

125-5d² = 120

-5d² = 120-125

-5d² = -5

d² = -5/-5

d² = 1

d = ±1

Now, if a=5 and d=−1 then the first three terms of the A.P are:

a−d=5−(−1)=5+1=6

a=5 and

a+d=5-1=4

And if a=5 and d=1 then the first three terms of the A.P are:

a−d=5−1=4

a=5 and

a+d= 5+1=6

So three terms are 6,5,4 Or 4,5,6

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