# The sum $ {S_n} $ of the terms of an arithmetic progression is defined by the formula $ {S_n} = 4{n^2} - 3n $ for any n. Write the first three terms of the progression.

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**Hint**: In the given problem, we are given the expression for the sum of n terms of an arithmetic progression. We are required to find the first three terms of the progression on the basis of already available information. For finding out the sum of an arithmetic progression, we need to know the first term, the common difference and the number of terms in the arithmetic progression. We can find out the common difference of an arithmetic progression by knowing the difference of any two consecutive terms of the series.

**:**

__Complete step-by-step answer__So, we have the sum of an arithmetic progression as $ {S_n} = 4{n^2} - 3n $

Now, we know that we can find the sum of the given arithmetic progression using the formula $ S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $ , where n is the number of terms in the AP, d is the common difference between any two consecutive terms of arithmetic progression and a is the first term of the series.

So, we can equate the formula of sum of n terms with the expression provided to us in the problem. So, we get, $ S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = 4{n^2} - 3n $

Now, we can take n common from the bracket. So, we get,

$ \Rightarrow \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = n\left( {4n - 3} \right) $

Now, taking $ \left( {\dfrac{1}{2}} \right) $ common from the bracket in the right side of the equation in order to resemble the left side of the equation, we get,

$ \Rightarrow \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = \dfrac{n}{2}\left( {8n - 6} \right) $

Now, comparing both the sides of the equation, we get,

$ \Rightarrow 2a + \left( {n - 1} \right)d = 8n - 6 $

Simplify the equation, we get,

$ \Rightarrow 2a + \left( {n - 1} \right)d = 8\left( {n - 1} \right) + 8 - 6 $

$ \Rightarrow 2a + \left( {n - 1} \right)d = 8\left( {n - 1} \right) + 2 $

So, comparing both the sides of the equation, we get,

$ 2a = 2 $ and $ d = 8 $

$ \Rightarrow a = 1 $ and $ d = 8 $

So, we get the first term of the arithmetic progression as $ 1 $ and the common difference of AP as $ 8 $ .

Now, we can find the first three terms of the progression using the general term for nth term of an AP.

So, we have, $ {a_n} = a + \left( {n - 1} \right)d $

So, the first term of the progression is $ {a_1} = a + \left( {1 - 1} \right)d = a = 1 $ .

Second term of the progression is $ {a_2} = a + \left( {2 - 1} \right)d = a + d = 1 + 8 = 9 $

Third term of the progression is $ {a_3} = a + \left( {3 - 1} \right)d = a + 2d = 1 + 2\left( 8 \right) = 17 $

So, the first three terms of the arithmetic progression are: $ 1 $ , $ 9 $ and $ 17 $ .

**So, the correct answer is “ $ 1 $ , $ 9 $ and $ 17 $ ”.**

**Note**: Arithmetic progression is a series where any two consecutive terms have the same difference between them. The common difference of an arithmetic series can be calculated by subtraction of any two consecutive terms of the series. The sum of n terms of an arithmetic progression can be calculated if we know the first term, the number of terms and difference of the arithmetic series as: $ S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] $ .

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