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प्रश्न
The sum of the first n terms of an AP is given by `s_n = ( 3n^2 - n) ` Find its
(i) nth term,
(ii) first term and
(iii) common difference.
उत्तर
Given : `s_n = ( 3n^2 - n) ` .............(1)
Replacing n by (n - 1) in (i), we get:
`s_(n-1) = 3 (n-1) ^2 - (n-1)`
= 3 (n2 - 2n +1 ) - n + 1
= 3n2 -7 n + 4
(i) Now , `T_n = ( s_n - s_( n-1))`
`=(3n^2 - n) - ( 3n^2 - 7n +4) = 6n -4 `
∴ nth term , Tn = (6n -4) ..................(ii)
(ii) Putting n = 1 in (ii), we get:
`T_1 = (6 xx 1) -4 = 2`
(iii) Putting n = 2 in (ii), we get:
`T_2 = (6xx 2 )-4 = 8`
∴ Common difference, d = T2 - T1 = 8 - 2 = 6
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