Question

The sum of the latter half of 2n terms of any A.P. is one-third the sum of the 3n terms of the same A.P.
A. True
B. False

Answer 1

Hint: The given question is related to arithmetic progression. Use the formulae for the nth term of an A.P. and the sum of nth term of an A.P. $S = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ and use it to find the sum of the latter half of \[2n\] terms of any A.P. by substituting \[2n\] in place of \[n\] and further obtain by subtracting the sum of n terms from the sum of total \[2n\] terms of that A.P. Simplify it to bring it in terms of the sum of the 3n terms of an A.P., which is given by \[{S_{3n}} = \dfrac{1}{3}\left( {\dfrac{{3n}}{2}\left( {2a + \left( {3n - 1} \right)d} \right)} \right)\].

Complete step by step answer:

Let us start by thinking of the sum formula of an A.P. and the fact that the sum of the latter half of 2n terms of any A.P. means that we need to find the sum of last n terms of that A.P.
We know by the sum formula of an A.P. that the sum of n terms is given by;
$S = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Thus, the sum of n terms from the last can be given by subtracting the sum of n terms from the sum of total 2n terms of that A.P.
\[ \Rightarrow {S_{{\text{req}}}} = \dfrac{{2n}}{2}\left( {2a + \left( {2n - 1} \right)d} \right) - \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\] ……(1)
Our objective is to show that the value of the expression in equation (1) is equal to one-third the sum of the 3n terms of the same A.P. which is given by;
\[{S_{3n}} = \dfrac{1}{3}\left( {\dfrac{{3n}}{2}\left( {2a + \left( {3n - 1} \right)d} \right)} \right)\] …….(2)
We now simplify the obtained expression to get the relation between the sum of the latter half of 2n terms of any A.P. and the sum of the 3n terms of the same A.P.
\[
   \Rightarrow {S_{{\text{req}}}} = \dfrac{n}{2}\left( {4a + 2\left( {2n - 1} \right)d - 2a - \left( {n - 1} \right)d} \right) \\
   \Rightarrow {S_{{\text{req}}}} = \dfrac{n}{2}\left( {2a + 2\left( {2n - 1} \right)d - \left( {n - 1} \right)d} \right) \\
   \Rightarrow {S_{{\text{req}}}} = \dfrac{n}{2}\left( {2a + d\left( {4n - 2 - n + 1} \right)} \right) \\
   \Rightarrow {S_{{\text{req}}}} = \dfrac{n}{2}\left( {\left( {2a + \left( {3n - 1} \right)d} \right)} \right) \\
 \]
Multiply the numerator and denominator by 3 on the right side we get;
\[ \Rightarrow {S_{{\text{req}}}} = \dfrac{1}{3}\left( {\dfrac{{3n}}{2}\left( {2a + \left( {3n - 1} \right)d} \right)} \right)\] …….(3)
Hence, by (1), (2), and (3); it can be concluded that the sum of the latter half of 2n terms of any A.P. is one-third of the 3n terms of the same A.P.
Hence, option (a) is the correct option.

Note: In the given question, the fact that the sum of the latter half of 2n terms of any A.P. means that we need to find the sum of last n terms of that A.P. Once using that condition, the sum is obtained, all you need to do is simplify and see if it is equal to the sum of 3n terms. Remember the basic formulas like the sum of first \[n\] terms and as we have formed the equations, do make the equations to get the desired result. Do the simplification of the expression properly.