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

1672 Points
```							hello babita. we know thattn= Sn – Sn-1so tn= 4n^2-3n – [4(n-1)^2-3(n-1)]= 4(2n-1)-3tn= 8n – 7so tn – tn-1= 8 which is the common difference, hence the sequence is an AP
```
10 months ago
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• 101 Video Lectures
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions