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Hint: In Arithmetic Progressions, we use the formula of sum of n terms i.e. ${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$. We will use this formula in this question as well. Another formula that we will use in the solution is i.e. ${a_n} = a + \left( {n - 1} \right)d$.

__Complete step-by-step answer:__

As we know, the sum of n terms is calculated by${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$ , we will put the value of n as 10 because the value of ${S_{10}}$ is already given to us by the question, we get-

$ \to {S_{10}} = \dfrac{{10}}{2}\left[ {a + {a_{10}}} \right]$

Substituting the value of ${S_{10}}$in the above equation, we get-

$

\to 125 = \dfrac{{10}}{2}\left[ {a + {a_{10}}} \right] \\

\\

\to 125 = 5\left[ {a + {a_{10}}} \right] \\

\\

\to 25 = a + {a_{10}} \\

$

Applying the formula ${a_n} = a + \left( {n - 1} \right)d$, in the above equation, we get-

$

\to 25 = a + a + 9d \\

\\

\to 25 = 2a + 9d \\

$

Let the equation $25 = 2a + 9d$ be equation 1-

$ \to 25 = 2a + 9d$ (equation 1)

It is given in the question that ${a_3} = 15$, so we will have-

$ \to a + 2d = 15$

Multiplying the equation 2 by 2, we get-

$ \to 2a + 4d = 30$ (equation 2)

Subtracting equation 1 from equation 2-

$ \Rightarrow - 5d = 5$

So, the value of d which we get from the above equation is $d = - 1$.

Putting the value of d into equation 1-

$

\to 25 = 2a - 9 \\

\\

\to 35 = 2a \\

\\

\to a = 17 \\

$

So, we find ${a_{10}}$ by the formula ${a_n} = a + \left( {n - 1} \right)d$, where $n = 10$, and putting the value of d found above, we get-

$

\to {a_{10}} = 17 - 9 \\

\\

\to {a_{10}} = 8 \\

$

Hence, the value of $d$ is -1 and the value of ${a_{10}}$ is 8.

Note: Remember the formula of sum of n terms i.e. ${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$ as it is used in most of the Arithmetic Progressions questions. Use the values given in the question wisely to find out the values asked by the question.

As we know, the sum of n terms is calculated by${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$ , we will put the value of n as 10 because the value of ${S_{10}}$ is already given to us by the question, we get-

$ \to {S_{10}} = \dfrac{{10}}{2}\left[ {a + {a_{10}}} \right]$

Substituting the value of ${S_{10}}$in the above equation, we get-

$

\to 125 = \dfrac{{10}}{2}\left[ {a + {a_{10}}} \right] \\

\\

\to 125 = 5\left[ {a + {a_{10}}} \right] \\

\\

\to 25 = a + {a_{10}} \\

$

Applying the formula ${a_n} = a + \left( {n - 1} \right)d$, in the above equation, we get-

$

\to 25 = a + a + 9d \\

\\

\to 25 = 2a + 9d \\

$

Let the equation $25 = 2a + 9d$ be equation 1-

$ \to 25 = 2a + 9d$ (equation 1)

It is given in the question that ${a_3} = 15$, so we will have-

$ \to a + 2d = 15$

Multiplying the equation 2 by 2, we get-

$ \to 2a + 4d = 30$ (equation 2)

Subtracting equation 1 from equation 2-

$ \Rightarrow - 5d = 5$

So, the value of d which we get from the above equation is $d = - 1$.

Putting the value of d into equation 1-

$

\to 25 = 2a - 9 \\

\\

\to 35 = 2a \\

\\

\to a = 17 \\

$

So, we find ${a_{10}}$ by the formula ${a_n} = a + \left( {n - 1} \right)d$, where $n = 10$, and putting the value of d found above, we get-

$

\to {a_{10}} = 17 - 9 \\

\\

\to {a_{10}} = 8 \\

$

Hence, the value of $d$ is -1 and the value of ${a_{10}}$ is 8.

Note: Remember the formula of sum of n terms i.e. ${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$ as it is used in most of the Arithmetic Progressions questions. Use the values given in the question wisely to find out the values asked by the question.

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