Question

The sum of the latter half of first \[200\] terms of an A.P. whose first term and common difference both are positive is equal to one-third of the sum of the first \[k\] terms of the same A.P., then the value of \[k\] is
(A) \[100\]
(B) \[250\]
(C) \[300\]
(D) \[350\]

Answer 1

Hint: The given question is related to the arithmetic progression. First, we use the formula for the latter half of \[2n\] terms of an A.P. and calculate the sum of the latter half of first \[200\] terms of an A.P. After that using formula for the sum of \[n\] terms of an A.P. , we find the sum of the first \[k\] terms of the A.P. then we equate the sum of the later half of first \[200\] terms to one-third of the sum of the first \[k\] terms of the A.P. and find the value of \[k\]
Formulas used:
(i) sum of the latter half of \[2n\] terms:
\[S = \dfrac{n}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]\]
(ii) sum of n terms of an A.P.:
\[S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]

Complete step-by-step solution:
First of all, let the first term of an A.P. be \[a\] and the common difference be \[d\]
Now using formula for sum of the latter half of \[2n\] terms
i.e., \[S = \dfrac{n}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]\]
In the question we have given the sum of the latter half of \[200\] terms
It means here, \[2n = 200\]
\[\therefore \] sum of the latter half of \[200\] terms will be
\[{S_{200}} = \dfrac{{100}}{2}\left[ {2a + \left( {3\left( {100} \right) - 1} \right)d} \right]\]
\[ = \dfrac{{100}}{2}\left[ {2a + \left( {299} \right)d} \right]{\text{ }} - - - \left( 1 \right)\]
Now we know that sum of n terms of an A.P. is equals to
\[S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
So, the sum of the first \[k\] terms of the A.P. will be equals to
\[{S_k} = \dfrac{k}{2}\left[ {2a + \left( {k - 1} \right)d} \right]{\text{ }} - - - \left( 2 \right)\]
Now, it is given that the sum of the latter half of first \[200\] terms of an A.P. is equal to one-third of the sum of the first \[k\] terms of the same A.P.
So, using equation \[\left( 1 \right)\] and \[\left( 2 \right)\] we can write,
\[\dfrac{{100}}{2}\left[ {2a + \left( {299} \right)d} \right]{\text{ }} = {\text{ }}\dfrac{1}{3}\left\{ {\dfrac{k}{2}\left[ {2a + \left( {k - 1} \right)d} \right]} \right\}\]
On cancelling \[2\] both sides from the denominator, we get
\[100\left[ {2a + \left( {299} \right)d} \right]{\text{ }} = {\text{ }}\dfrac{k}{3}\left[ {2a + \left( {k - 1} \right)d} \right]\]
\[ \Rightarrow 2a\left( {\dfrac{k}{3} - 100} \right) + \dfrac{{k\left( {k - 1} \right)d}}{3} - 29900d = 0\]
On simplifying it, we get
\[ \Rightarrow 2a\left( {\dfrac{{k - 300}}{3}} \right) = \left( {29900 - \dfrac{{k\left( {k - 1} \right)}}{3}} \right)d\]
On cancelling \[3\] on both sides, we get
\[2a\left( {k - 300} \right){\text{ }} = {\text{ }}\left( {89700 - {k^2} + k} \right)d{\text{ }} - - - \left( 3 \right)\]
\[ \Rightarrow \dfrac{a}{d} = \dfrac{{\left( {89700 - {k^2} + k} \right)}}{{2k - 600}}\]
Now, it is given that the first term and the common difference both are positive.
\[\therefore \dfrac{a}{d} \geqslant 0\]
\[ \Rightarrow \dfrac{a}{d} = \dfrac{{\left( {89700 - {k^2} + k} \right)}}{{2k - 600}}{\text{ }} \geqslant 0\]
So, either \[89700 - {k^2} + k{\text{ }} \geqslant 0{\text{ }}\]
\[ \Rightarrow {k^2} - k - 89700{\text{ }} \leqslant 0{\text{ }}\] which is not possible as \[\left( {{k^2} - k - 89700} \right)\] is coefficient of \[d\] and it is already given that \[d\] is positive.
Or \[2k - 600{\text{ }} \geqslant 0\]
\[ \Rightarrow k \geqslant 300\]
\[ \Rightarrow k = 300\] which is the required answer.
Hence, option (C) is the correct answer.

Note: While solving this question, it can also be solved by using one important defined rule which is the sum of the later half of \[2n\] terms of any A.P. is one-third the sum of the \[3n\] terms of the same A.P.
So according to question, the sum of the latter half of \[200\] terms of any A.P. will be one-third the sum of the \[300\] terms of the same A.P.
And it is given that the sum of the latter half of \[200\] terms of any A.P. is one-third the sum of the \[k\] terms of the same A.P.
Hence on comparing both the condition stated above, we get
\[k = 300\] which is the required answer.