Question

if an= n square -1 find a(n-1) and a n+1​

Answer 1

Answer:

Step-by-step explanation:

Answer : N ^2 = 23164969

Solution : Let N be a positive number containing ' m ' digits in it such that N is between N1 and N2 where N1=25*10^(m-2) and N2= 50*10^(m-2) . Then

N ^2 = (N-N1)*10^m +(N2 - N )^2.

Here , m=4, N1=2500 , N2 = 5000 , N = 4813

N ^2 = (4813 -2500 ) *10^4 + (5000 - 4813 )^2

= 23130000 + 187 ^2

= 23130000 + 34969

= 23164969

Answer 2

Answer:

a(n-1)= - 3

a(n+1)= 3

Step-by-step explanation:

a1=1-1

=0

a2=4-1

=3

d=a2-a1

=3-0

=3

AP=a(n-1), an, a(n+1)

Therefore AP= -3, 0, 3

Hope it helps you

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