Question

If p times the \[{{p}^{th}}\] term of an A.P. is equal to q times the \[{{q}^{th}}\] term of an A.P. Then \[{{\left( p+q \right)}^{th}}\] term is
(a) 0
(b) 1
(c) 2
(d) 3

Answer 1

Hint: Use the general \[{{m}^{th}}\] term for an AP \[{{t}_{m}}=a+\left( m-1 \right)d\] where ‘a’ is the first term of the AP and ‘d’ is its common difference, to formulate the two conditions given in the equations. Now, use this to calculate \[{{t}_{n}}=\left( p+q-n \right)\].
We know the general \[{{m}^{th}}\] term for an AP
\[{{t}_{m}}=a+\left( m-1 \right)d\]

Complete step by step answer:
According to the question it is asked to us to find the term \[{{\left( p+q \right)}^{th}}\] if p times of \[{{p}^{th}}\] term of an A.P. is equal to q times of \[{{q}^{th}}\] term.
Since, we know that the general \[{{m}^{th}}\] term of AP is formed with the help of the formula.
\[{{a}_{n}}=a+\left( n-1 \right)d\]
So, for the \[{{p}^{th}}\] term, \[{{a}_{p}}=a+\left( p-1 \right)d\]
And the \[{{q}^{th}}\] term, \[{{a}_{q}}=a+\left( q-1 \right)d\]
And \[{{\left( p+q \right)}^{th}}\] term, \[{{a}_{p+q}}=a+\left( p+q-1 \right)d\]
If we solve to these, then
\[p\left( a+\left( p-1 \right)d \right)=q\left( a+\left( q-1 \right)d \right)\]
If we take R.H.S. to the R.H.S. then,
\[\begin{align}
  & p\left( a+\left( p-1 \right)d \right)-q\left( a+\left( q-1 \right)d \right)=0 \\
 & a\left( p-q \right)+\left( {{p}^{2}}-{{q}^{2}} \right)d+\left( q-p \right)d=0 \\
\end{align}\]
\[a\left( p-q \right)+\left( p-q \right)\left( p+q \right)d\left( q-p \right)=0\]
\[a\left( p-q \right)+\left( p-q \right)d+\left( p+q-1 \right)d=0\]
Now cancel \[\left( p-q \right)\] from both the term
\[a+\left( p+q-1 \right)d=0\]
which is equal to \[{{\left( p+q \right)}^{th}}\] term. And this is equal to zero.

So, the correct answer is “Option a”.

Note: While solving these types of questions you have to use the formula \[{{a}_{n}}=a+\left( n-1 \right)d\] because this is the exact formula which provide the exact value of \[{{n}^{th}}\] term in A.P. And if we compare any two terms of an A.P. then we can find the all required details of an A.P.