# What is the first term of arithmetic progression  if   `S_n=3n^2-n` and `d=6`  ? 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`...

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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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