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Grade: 11
        
Sn=4n^2-3n find tn and show that the sequence is an A.P
9 months ago

Answers : (1)

Aditya Gupta
1401 Points
							
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
9 months ago
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