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

Find the sum of the two middle most terms of the A.P: −43,−1,−23,−13,.....,413. - \dfrac{4}{3}, - 1, - \dfrac{2}{3}, - \dfrac{1}{3},.....,4\dfrac{1}{3}.

Explanation

Solution

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 an=a+(n−1)d{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. an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d

Complete step by step solution:
As, the given AP is −43,−1,−23,−13,.....,413. - \dfrac{4}{3}, - 1, - \dfrac{2}{3}, - \dfrac{1}{3},.....,4\dfrac{1}{3}.
So, the first term a=−43a = - \dfrac{4}{3}
And Common difference d=−1−(−43)d{\rm{ }} = {\rm{ }} - 1 - \left( {\dfrac{{ - 4}}{3}} \right){\rm{ }}
On opening the brackets we get,
d=−1+43d{\rm{ }} = {\rm{ }} - 1 + \dfrac{4}{3}{\rm{ }}
By taking L.C.M.
d=−3+43d{\rm{ }} = {\rm{ }}\dfrac{{ - 3 + 4}}{3}{\rm{ }}
On further simplification: d=13d{\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 an=413{a_n} = 4\dfrac{1}{3}
Hence on converting mixed fraction into proper fraction we get, an=133{a_n} = \dfrac{{13}}{3}
Thus, by using the formula of an A.P. which is an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d
Substituting all the values of an{a_n} , a and d . So, that we calculate the value of n.
133=−43+(n−1)13\dfrac{{13}}{3} = - \dfrac{4}{3} + \left( {n - 1} \right)\dfrac{1}{3}
On taking −43 - \dfrac{4}{3} on left side we get,
133+43=(n−1)13\dfrac{{13}}{3} + \dfrac{4}{3} = \left( {n - 1} \right)\dfrac{1}{3}
On simplifying:
13+43=(n−1)13\dfrac{{13 + 4}}{3} = \left( {n - 1} \right)\dfrac{1}{3}
173=(n−1)13\dfrac{{17}}{3} = \left( {n - 1} \right)\dfrac{1}{3}
Cancelling 3 from denominator from both sides,
We get, 17=(n−1)17 = \left( {n - 1} \right)
Hence, n come out to be 18 i.e. n=18n = 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 (182)th\left( {\dfrac{{18}}{2}} \right)th term and (182+1)th\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+(n1−1)da + \left( {{n_1} - 1} \right)d + a+(n2−1)da + \left( {{n_2} - 1} \right)d where n1=9{n_1} = 9 and n2=10{n_2} = 10
So, on substituting the values we get: −43+(9−1)13 - \dfrac{4}{3} + \left( {9 - 1} \right)\dfrac{1}{3} + (−43+(10−1)13)\left( { - \dfrac{4}{3} + \left( {10 - 1} \right)\dfrac{1}{3}} \right)
On solving we get, −43+83−43+3⇒3 - \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. (n2,n2+1)\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.