Question

Find the sum of the two middle most terms of the A.P: \[ - \dfrac{4}{3}, - 1, - \dfrac{2}{3}, - \dfrac{1}{3},.....,4\dfrac{1}{3}.\]

Answer 1

Hint:
According to the question, first calculate the first term, common difference and the last terms for the calculation of the number of terms using the formula \[{a_n} = a + \left( {n - 1} \right)d\]. Then, calculate the sum of the two middle most terms if the A.P.

Formula used:
Here, we use the formula of sum of n terms i.e. \[{a_n} = a + \left( {n - 1} \right)d\]

Complete step by step solution:
As, the given AP is \[ - \dfrac{4}{3}, - 1, - \dfrac{2}{3}, - \dfrac{1}{3},.....,4\dfrac{1}{3}.\]
So, the first term \[a = - \dfrac{4}{3}\]
And Common difference \[d{\rm{ }} = {\rm{ }} - 1 - \left( {\dfrac{{ - 4}}{3}} \right){\rm{ }}\]
On opening the brackets we get,
\[d{\rm{ }} = {\rm{ }} - 1 + \dfrac{4}{3}{\rm{ }}\]
By taking L.C.M.
\[d{\rm{ }} = {\rm{ }}\dfrac{{ - 3 + 4}}{3}{\rm{ }}\]
On further simplification: \[d{\rm{ }} = {\rm{ }}\dfrac{1}{3}{\rm{ }}\]
Let us suppose there are n terms in the given AP.
Therefore, last term of an A.P that is \[{a_n} = 4\dfrac{1}{3}\]
Hence on converting mixed fraction into proper fraction we get, \[{a_n} = \dfrac{{13}}{3}\]
Thus, by using the formula of an A.P. which is \[{a_n} = a + \left( {n - 1} \right)d\]
Substituting all the values of \[{a_n}\] , a and d . So, that we calculate the value of n.
\[\dfrac{{13}}{3} = - \dfrac{4}{3} + \left( {n - 1} \right)\dfrac{1}{3}\]
On taking \[ - \dfrac{4}{3}\] on left side we get,
\[\dfrac{{13}}{3} + \dfrac{4}{3} = \left( {n - 1} \right)\dfrac{1}{3}\]
On simplifying:
\[\dfrac{{13 + 4}}{3} = \left( {n - 1} \right)\dfrac{1}{3}\]
\[\dfrac{{17}}{3} = \left( {n - 1} \right)\dfrac{1}{3}\]
Cancelling 3 from denominator from both sides,
We get, \[17 = \left( {n - 1} \right)\]
Hence, n come out to be 18 i.e. \[n = 18\]
So, the given AP consists of 18 terms. hence, there are two middle terms in the given AP. The middle terms of the given AP are \[\left( {\dfrac{{18}}{2}} \right)th\] term and \[\left( {\dfrac{{18}}{2} + 1} \right)th\] term which are 9th term and 10th term in the given A.P.
So, The Sum of the middle most terms of the given AP is: 9th term + 10th term which is further equals to \[a + \left( {{n_1} - 1} \right)d\] + \[a + \left( {{n_2} - 1} \right)d\] where \[{n_1} = 9\] and \[{n_2} = 10\]
So, on substituting the values we get: \[ - \dfrac{4}{3} + \left( {9 - 1} \right)\dfrac{1}{3}\] + \[\left( { - \dfrac{4}{3} + \left( {10 - 1} \right)\dfrac{1}{3}} \right)\]
On solving we get, \[ - \dfrac{4}{3} + \dfrac{8}{3} - \dfrac{4}{3} + 3 \Rightarrow 3\]

Hence, the sum of the middle most terms of the given AP is 3.

Note:
To solve these types of questions, you should keep in mind that we should calculate n using the formula of an A.P series. Then you should try to find out the middle most terms i.e. \[\left( {\dfrac{n}{2},\dfrac{n}{2} + 1} \right)\] with the help of n. Hence, to find their sum we can use the formula of an A.P series.