What is the first term of arithmetic progression if `S_n=3n^2-n` and `d=6` ?
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Note that in arithmetic progression, n refers to the number of terms. If we set n=2, the sum of the two terms (`a_1 ` and `a_2` ) is:
`S_n = 3n^2 - n`
`S_n = 3(2)^2-2`
`S_n=10`
Replacing `S_n` with `a_1` + `a_2` , yields:
`a_1 + a_2 = 10` (Let this be EQ1.)
Also, the d refers to the difference between consecutive terms in arithmetic series. So, the difference between `a_1` and `a_2` is:
`a_2- a_1 = d`
`a_2-a_1=6 ` (Let this be EQ2.)
To solve for the first term `a_1` , subtract EQ2 from EQ1.
`a_1+a_2=10 ` `a_1 + a_2=10`
`(-)` `a_2-a_1=6` ===> `(-)` `-a_1 +a_2=6`
` ------------`
`2a_1+ 0=4`
Then, isolate `a_1` .
`2a_1 = 4`
`a_1=4/2`
`a_1=2`
Hence, the first term in the arithmetic progression is 2.
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