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**Hint:**We will find the common difference of the series by taking any two consecutive terms let us say ${{\text{r}}^{{\text{th}}}}$ and ${(r - 1)^{th}}$ terms i.e., we’ll find the value of ${a_r} - {a_{r - 1}}$, if it comes out to be constant the given series will form an A.P. as an A.P. have a constant common difference throughout the series.

To find the sum of the first 20 terms of the series we’ll use the formula of the sum of first n terms of an A.P.

**Complete step by step solution:**

Given data: ${a_n} = 3 - 4n,$

Let a random term and its preceding term to find the common difference of the series

Let's take ${{\text{r}}^{{\text{th}}}}$ and ${(r - 1)^{th}}$ terms

$

{{\text{a}}_{\text{r}}}{\text{ = 3 - 4r}} \\

\Rightarrow {{\text{a}}_{{\text{r - 1}}}}{\text{ = 3 - 4(r - 1)}} \\

\Rightarrow {{\text{a}}_{{\text{r - 1}}}}{\text{ = 3 - 4r + 4}} \\

\Rightarrow {{\text{a}}_{{\text{r - 1}}}}{\text{ = 7 - 4r}} \\

$

Now common difference say d,

\[

{\text{d = }}{{\text{a}}_{\text{r}}}{\text{ - }}{{\text{a}}_{{\text{r - 1}}}} \\

\Rightarrow {\text{d = 3 - 4r - (7 - 4r)}} \\

\Rightarrow {\text{d = 3 - 4r - 7 + 4r}} \\

\Rightarrow {\text{d = - 4}} \\

\]

Putting the values of ${a_r}$ and ${a_{r - 1}}$

Since the common difference(d) is constant, it can be said that ${a_1},{a_2},{a_3}.......$ will form an A.P.

Using ${a_n} = 3 - 4n,$

Putting \[n = 1\],

$

{{\text{a}}_{\text{1}}}{\text{ = 3 - 4(1)}} \\

\therefore {{\text{a}}_{\text{1}}}{\text{ = - 1}} \\

$

Putting \[n = 2\],

$

{{\text{a}}_{\text{2}}}{\text{ = 3 - 4(2)}} \\

\therefore {{\text{a}}_{\text{2}}}{\text{ = - 5}} \\

$

Putting \[n = 3\],

$

{a_3} = 3 - 4(3) \\

\therefore {a_3} = - 9 \\

$

Therefore, the A.P. will be -1,-5,-9……….

Now, Sum of first n terms of an A.P. i.e. ${S_n}$is given by,

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

Using this formula, we get,

${S_{20}} = \dfrac{{20}}{2}\left[ {2{a_1} + \left( {20 - 1} \right)( - 4)} \right]$

Using ${a_n} = 3 - 4n,$we get,

$

= 10\left[ {2(3 - 4(1)) + 19( - 4)} \right] \\

= 10\left[ {2(3 - 4) - 76} \right] \\

= 10\left[ {2( - 1) - 76} \right] \\

= 10\left[ { - 2 - 76} \right] \\

= 10\left[ { - 78} \right] \\

= - 780 \\

$

Therefore,

${S_{20}} = - 780$

**Hence, the AP is -1,-5,-9,…….. and ${S_{20}} = - 780$**

**Note:**We can also find the value of ${S_{20}}$ using the alternative formula for the sum of first n terms of an A.P. can also be written as

${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$

Using ${a_n} = 3 - 4n,$we get

\[

{S_{20}} = \dfrac{{20}}{2}\left[ {{a_1} + {a_{20}}} \right] \\

= \dfrac{{20}}{2}\left[ {3 - 4(1) + 3 - 4(20)} \right] \\

= 10\left[ {3 - 4 + 3 - 80} \right] \\

= 10\left[ {6 - 84} \right] \\

= 10\left[ { - 78} \right] \\

= - 780 \\

\]

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