Question

If an AP consists of \[n\] terms and the sum of the first three is \[X\] and the sum of the last three terms is \[Y\] then show that sum of all terms is equal to \[\left( {\dfrac{n}{6}} \right)\left( {X + Y} \right)\] ?

Answer 1

Hint: Here we are asked to show that the sum of all the terms in the given AP is equal to \[\left( {\dfrac{n}{6}} \right)\left( {X + Y} \right)\]. First, assume the sum of the first three terms \[X\] and the sum of the last three terms \[Y\] on the arithmetic progression method. Then used, to sum \[n\] up, terms formulae. We used the difference value to find three \[{S_n}\] values. Then sum all \[{S_n}\] values and apply the \[X\] and \[Y\] in the sum of all \[{S_n}\] values. Finally, we prove that answer.

Formulae used:
The sum of all \[n\] terms in the arithmetic progression formulae is
\[{S_n} = \dfrac{n}{2}\left( {{a_1} + {a_n}} \right)\]
\[{a_1}\] is denoted by the first term and \[{a_n}\] is denoted by the last term.

Complete step-by-step answer:
Given that arithmetic progression consists of \[n\] terms.
The Sum of the first three terms is \[X\] and the sum of the last three terms is \[Y\].
assume that the first three terms of an arithmetic progression \[{a_1},\,{a_{2,\,}}{a_3}\] are.
So that, \[{a_1} + {a_2} + {a_3} = X\]
And assume that the last three terms of an arithmetic progression \[{a_n},\,{a_{n - 1,\,}}{a_{n - 2}}\] are.
So that, \[{a_n} + {a_{n - 1}} + {a_{n - 2}} = Y\]
The sum of all \[n\] terms in the arithmetic progression formulae is
\[{S_n} = \dfrac{n}{2}\left( {{a_1} + {a_n}} \right)\]
\[{a_1}\] is denoted by the first term and \[{a_n}\] is denoted by the last term.
Now our first term and last term also \[{a_1}\] and \[{a_n}\].
So that,
\[{S_n} = \dfrac{n}{2}\left( {{a_1} + {a_n}} \right)\]
Consider this as the first \[{S_n}\] value.
In arithmetic progression \[d\] is denoted by the difference value of two terms.
So that we add \[d\] value in the first term we get the second value and subtract \[d\] the value in the last term we get the previous term's previous value.
It means
\[{a_1} + d = {a_2}\]
\[{a_n} - d = {a_{n - 1}}\]
Now we find \[{S_n}\] value
\[{S_n} = \dfrac{n}{2}\left( {{a_2} + {a_{n - 1}}} \right)\]
Consider this as the second \[{S_n}\] value.
Similarly,
Add \[d\] value to the second value, get the third value and subtract \[d\] value to the last previous term so we get another previous value.
It means that,
\[{a_2} + d = {a_3}\]
\[{a_{n - 1}} + d = {a_{n - 2}}\]
Now we find \[{S_n}\] value.
\[{S_n} = \dfrac{n}{2}\left( {{a_3} + {a_{n - 2}}} \right)\]
Consider this is third \[{S_n}\] values
Now add to all the \[{S_n}\] values
\[{S_n} + {S_n} + {S_n} = \dfrac{n}{2}\left( {{a_1} + {a_n}} \right) + \dfrac{n}{2}\left( {{a_2} + {a_{n - 1}}} \right) + \dfrac{n}{2}\left( {{a_3} + {a_{n - 2}}} \right)\]
Add left-hand \[{S_n}\] values and correctly arrange the right-hand side terms,
\[3{S_n} = \dfrac{n}{2}\left( {{a_1} + {a_2} + {a_3}} \right) + \dfrac{n}{2}\left( {{a_n} + {a_{n - 1}} + {a_{n - 2}}} \right)\]
And right-hand side take the common term on \[\dfrac{n}{2}\]
\[3{S_n} = \dfrac{n}{2}\left( {\left( {{a_1} + {a_2} + {a_3}} \right) + \left( {{a_n} + {a_{n - 1}} + {a_{n - 2}}} \right)} \right)\]
Now apply the sum of the first three terms value and the sum of the last three terms value.
\[3{S_n} = \dfrac{n}{2}\left( {X + Y} \right)\]
Now divide \[3\] into both sides
\[{S_n} = \dfrac{n}{6}\left( {X + Y} \right)\]
Now we get a sum of all terms \[ = \dfrac{n}{6}\left( {X + Y} \right)\].

Note: In this problem we first need to form an expression/equation from the given statement to find the sum of the first three and last three terms of the arithmetic progression. Then using this we found the sum of missing terms. We have to be more careful in splitting the sum of the terms. Finally, we have to simplify as much as we can to reach the required sum.