Question

The algebraic expression of a sequences is
xn = n square3 - 6n square2 + 13n - 7
Is it an arithmetic sequence?

Answer 1

Step-by-step explanation:

For it to be an AP, xn+1 – xn = constant

xn+1-xn = {(n+1)3-6(n+1)2+13(n+1)-7}-{ n3 – 6n2+ 13n – 7}

= (n+1)3-n3-6{(n+1)2-n2}+13(n+1-n)-7+7

= (n+1-n)(n2+2n+1+n2+n+n2)-6(2n+1)+13

= 1× (3n2+3n+1)-12n+7

= 3n2-9n+8

Which is not independent of n. Hence, the sequence is not an AP.