Question

If the \[p\] th term is of an AP be \[1/q\] and \[q\] th term be \[1/p\], then the sum of its \[pq\]th terms will be ?

Answer 1

Hint:The given question is based on the topic Arithmetic Progression (AP). An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. An AP is of the form \[a,a + d,a + 2d,a + 3d,...\] where \[a\]the first term is and \[d\] is the common difference. First find the value of \[a\] and \[d\]. Then by substituting the values in the sum of \[n\] terms formulas,\[pq\] th terms can be found.

Formulas used:
In an AP with first term \[a\] and common difference \[d\], the \[n\] th term (or the general term) is given by \[{a_n} = a + (n - 1)d\].
The sum of the first \[n\] terms of an AP is given by \[S = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]\].

Complete step by step answer:
Given that \[p\] th and \[q\] th term of an AP be \[1/q\] and \[1/p\] respectively.
Let us consider, \[a\] be the first term and \[d\] be the common difference of the AP.
We know that \[{a_n} = a + (n - 1)d\], let substitute the values in this formula.
For \[p\] th term, \[n = p\]
\[{a_p} = a + (p - 1)d\]
Its given that \[p\] th term is \[\dfrac{1}{q}\],
\[\dfrac{1}{q} = a + (p - 1)d\]
This can be rewrite us,
\[a + (p - 1)d = \dfrac{1}{q}\] ……………………. (\[1\])
Similarly, for \[q\] th, \[n = q\]
\[{a_q} = a + (q - 1)d\]
We know \[q\] th term \[ = \dfrac{1}{p}\]
\[\therefore \dfrac{1}{p} = a + (q - 1)d\]
This can also be written as,
\[a + (q - 1)d = \dfrac{1}{p}\] ………………….. (\[2\])
Now Subtracting equation (\[1\]) and (\[2\]),
\[{\underline a + (p - 1)d = \dfrac{1}{q} \\
a + (q - 1)d = \dfrac{1}{p} \\
_{( - )}} \\
0 + (p - 1)d - (q - 1)d = \dfrac{1}{q} - \dfrac{1}{p} \\ \]
Now taking \[d\] as common in L.H.S.,
\[\left( {(p - 1) - (q - 1)} \right)d = \dfrac{1}{q} - \dfrac{1}{p}\]
Now multiplying \[ - \] with \[(q - 1)\] we will get,
\[(p - 1 - q + 1)d = \dfrac{1}{q} - \dfrac{1}{p}\]
Here \[ - 1\] and \[ + 1\] will get cancel,
\[(p - q)d = \dfrac{1}{q} - \dfrac{1}{p}\]
Now by taking L.C.M. in R.H.S.,
L.C.M. is \[pq\], multiplying \[\dfrac{1}{q}\] by \[p\] and \[\dfrac{1}{p}\] by \[q\],
\[(p - q)d = \dfrac{{p - q}}{{pq}}\]
\[p - q\] in both the sides will get cancel,
\[d = \dfrac{1}{{pq}}\]
Now substituting the value of \[d = \dfrac{1}{{pq}}\] in either equation (\[1\]) or (\[2\]) we will get the value of \[a\].
Let us substitute \[d = \dfrac{1}{{pq}}\] in equation (\[1\]),
\[a + (p - 1)d = \dfrac{1}{q}\]
\[\Rightarrow a + (p - 1)\dfrac{1}{{pq}} = \dfrac{1}{q}\]
Multiplying \[\dfrac{1}{{pq}}\] with \[p - 1\]
\[a + \dfrac{p}{{pq}} - \dfrac{1}{{pq}} = \dfrac{1}{q}\]
Let us move the values in L.H.S to R.H.S, the \[ + \] will become \[ - \] and vice-versa.
\[a = \dfrac{1}{q} - \dfrac{p}{{pq}} + \dfrac{1}{{pq}}\]
By taking L.C.M.,
L.C.M. \[ = pq\], multiply \[\dfrac{1}{q}\] by \[p\]
\[a = \dfrac{{pq - pq + 1}}{{pq}}\]
\[pq\] and \[ - pq\] will get cancel,
\[a = \dfrac{1}{{pq}}\].
Hence we find the values of \[a = \dfrac{1}{{pq}}\] and \[d = \dfrac{1}{{pq}}\].
In our question we are asked to find the sum of \[pq\] th, we have the formula,
\[S = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]\], by substituting the values where \[n = pq\],
\[\Rightarrow S = \dfrac{{pq}}{2}\left[ {2\left( {\dfrac{1}{{pq}}} \right) + \left( {pq - 1} \right)\left( {\dfrac{1}{{pq}}} \right)} \right]\]
\[\Rightarrow S= \dfrac{{pq}}{2}\left[ {\dfrac{2}{{pq}} + \left( {pq - 1} \right)\left( {\dfrac{1}{{pq}}} \right)} \right]\]
Multiplying \[(pq - 1)\] by \[\dfrac{1}{{pq}}\] separately,
\[ \Rightarrow S=\dfrac{{pq}}{2}\left[ {\dfrac{2}{{pq}} + \dfrac{{pq}}{{pq}} - \dfrac{1}{{pq}}} \right]\]
There denominators are same, therefore,
\[\Rightarrow S= \dfrac{{pq}}{2}\left[ {\dfrac{{2 + pq - 1}}{{pq}}} \right]\]
Here, \[2 - 1 = 1\]
\[\Rightarrow S=\dfrac{{pq}}{2}\left[ {\dfrac{{1 + pq}}{{pq}}} \right]\]
Now, \[\dfrac{{pq}}{2}\] and \[\dfrac{{1 + pq}}{{pq}}\] are in multiplication, thus \[pq\] in numerator and denominator will get cancel,
\[\Rightarrow S=\dfrac{1}{2}\left[ {1 + pq} \right]\]
Multiplying \[1 + pq\] by \[\dfrac{1}{2}\] we will get,
\[\Rightarrow S=\dfrac{1}{2} + \dfrac{{pq}}{2}\]
\[\therefore S = \dfrac{1}{2} + \dfrac{{pq}}{2}\]
Here the denominators are the same thus \[S = \dfrac{{1 + pq}}{2}\].

Hence the sum of \[pq\] th term is \[\dfrac{{1 + pq}}{2}\].

Note: In this question we solved using the variables. One should be careful in taking L.C.M. for variables and the sign maths. If a sign goes to the opposite side of equal to then it will attain its opposite sign. For example, \[ + \] will become \[ - \] and vice-versa. \[ \times \] will become \[ \div \] and vice-versa. And just by substituting the values in the \[{a_n}\] and \[{S_n}\] formula we will get the answer.