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

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 ($2$),

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.