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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....

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

Explanation

Solution

Use the general mth{{m}^{th}} term for an AP tm=a+(m−1)d{{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 tn=(p+q−n){{t}_{n}}=\left( p+q-n \right).
We know the general mth{{m}^{th}} term for an AP
tm=a+(m−1)d{{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 (p+q)th{{\left( p+q \right)}^{th}} if p times of pth{{p}^{th}} term of an A.P. is equal to q times of qth{{q}^{th}} term.
Since, we know that the general mth{{m}^{th}} term of AP is formed with the help of the formula.
an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d
So, for the pth{{p}^{th}} term, ap=a+(p−1)d{{a}_{p}}=a+\left( p-1 \right)d
And the qth{{q}^{th}} term, aq=a+(q−1)d{{a}_{q}}=a+\left( q-1 \right)d
And (p+q)th{{\left( p+q \right)}^{th}} term, ap+q=a+(p+q−1)d{{a}_{p+q}}=a+\left( p+q-1 \right)d
If we solve to these, then
p(a+(p−1)d)=q(a+(q−1)d)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,

& 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{aligned}$$ $$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.