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प्रश्न
The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
विकल्प
24th term
27th term
26th term
25th term
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उत्तर
Here, the sum of first n terms is given by the expression,
`S_n = 3n^2 + 5n`
We ned to find which term of the A.P. is 164.
Let us take 164 as the nth term.
So we know that the nthterm of an A.P. is given by,
`a_n = S_n - S_( n-1) `
So,
`164 = S_n - S_( n-1) `
`164 = 3n^2 + 5n - [ 3(n-1)^2 + 5(n - 1 ) ]`
Using the property,
`( a - b)^2 = a^2 + n^2 - 2ab`
We get,
`164 = 3n^2 + 5n - [3 (n^2 + 1 - 2n ) + 5 (n -1)]`
`164 = 3n^2 + 5n - [ 3n^2 + 3 - 6n + 5n - 5 ]`
`164 = 3n^2 + 5n - (3n^2 - n - 2)`
`164 = 3n^2 + 5n - 3n^2 + n + 2 `
164 = 6n + 2
Further solving for n, we get
6n = 164 - 2
` n = 162/6`
n = 27
Therefore, 164 is the 27th term of the given A.P.
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