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Sn=4n^2-3n find tn and show that the sequence is an A.P

Sn=4n^2-3n find tn and show that the sequence is an A.P

Grade:11

1 Answers

Aditya Gupta
2081 Points
5 years ago
hello babita. we know that
tn= Sn – Sn-1
so tn= 4n^2-3n – [4(n-1)^2-3(n-1)]
= 4(2n-1)-3
tn= 8n – 7
so tn – tn-1= 8 which is the common difference, hence the sequence is an AP

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