Question

Is 0 a term of A.P 31,8,25

Answer 1

Here,

AP=31,8,25...

a=31

d=8-31

=-23

Let,

an=0

a+(n-1)d=0

31+(n-1)(-23)=0

(n-1)(-23)=-31

(n-1)=$\frac{ - 31}{ - 23}$

n=$\frac{31}{23} + 1$

n=$\frac{31 + 23}{23}$

n=$\frac{54}{23}$

But n should be positive integer.

So, 0 is not a term of AP

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Answer 2

Answer:

Step-by-step explanation:

31-3