The algebraic algebraic expression of a sequence is xn=n^3-6n^2+13n-7 is it an arithemetic sequence
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Answer:
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.
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