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In an AP, given ${a_3} = 15$, ${S_{10}} = 125$, find d and ${a_{10}}$.

Answer Verified Verified
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.