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**Hint:**Problems of this type can be easily solved by using the formulas of Arithmetic Progressions. We use the formula of the summation of the terms of an A.P to get an expression and equate it to $187$where we will get the value of the first term of the A.P. Upon doing that we equate the expression of the 17th term to $-13$ which will give us the value of common difference of the A.P, thus we determine the A.P.

**Complete step by step answer:**

We are given that the sum of first $17$ terms of the A.P is $187$and the ${{17}^{th}}$ term is $-13$ .

We use the formula of summation of A.P as shown below

Sum of $n$ terms of an A.P is $\dfrac{n}{2}\left( a+l \right)$

Here $a$ is the value of first term of the A.P and $l$ is the ${{n}^{th}}$ term of the A.P

We already know that $n=17$ and $l=-13$

Therefore, the expression of the summation becomes $\dfrac{17}{2}\left( a-13 \right)$

We now equate the above expression to the given value of sum i.e., $187$ as shown below

$\dfrac{17}{2}\left( a-13 \right)=187$

$\Rightarrow 17\left( a-13 \right)=187\times 2$

$\Rightarrow 17a-17\times 13=187\times 2$

$\Rightarrow 17a-221=374$

$\Rightarrow 17a=374+221$

$\Rightarrow 17a=595$

$\Rightarrow a=\dfrac{595}{17}$

$\therefore a=35$

Hence, the 1st term of the A.P is $35$

We also know that the formula of the value of ${{n}^{th}}$ term of the A.P is $a+\left( n-1 \right)d$

Here, $d$ is the common difference between two consecutive terms of the A.P

We can equate the above expression to $-13$ by putting $a=35$ and $n=17$ as shown below

$35+\left( 17-1 \right)d=-13$

$\Rightarrow \left( 17-1 \right)d=-13-35$

$\Rightarrow 16d=-48$

$\Rightarrow d=-\dfrac{48}{16}$

$\therefore d=-3$

Hence, the common difference between two consecutive terms of the A.P is $-3$

Therefore, the A.P is \[35,\text{ }32,\text{ }29,\text{ }26,\text{ }23,\text{ }20,\text{ }17,\text{ }14,\text{ }11,\text{ }8,\text{ }5,\text{ }2,\text{ }-1,\text{ }-4,\text{ }-7,\text{ }-10,\text{ }-13\]

**Note:**While developing the expression of summation of the terms of A.P we could have also used the formula for summation $\dfrac{n}{2}\left\{ 2a+\left( n-1 \right)d \right\}$ but the problem is the solution will become lengthier. This happens due to the fact that we have to solve two equations simultaneously to get the two variables $a$ and $d$ rather than simply solving each of the equations one by one to get the solutions.

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