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प्रश्न
Find the sum:
`4 - 1/n + 4 - 2/n + 4 - 3/n + ...` upto n terms
उत्तर
Here, first term, a = `4 - 1/n`
Common difference,
d = `(4 - 2/n) - (4 - 1/n)`
= `(-2)/n + 1/n`
= `(-1)/n`
∵ Sum of n terms of an AP,
Sn = `n/2[2a + (n - 1)d]`
⇒ Sn = `n/2[2(4 - 1/n) + (n - 1)((-1)/n)]`
= `n/2{8 - 2/n - 1 + 1/n}`
= `n/2(7 - 1/n)`
= `n/2 xx ((7n - 1)/n)`
= `(7n - 1)/2`
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