Solveeit Logo
Login

Question

Question: If \(m\) times the \({m^{th}}\) term of an A.P. is equal to \(n\) times its \({n^{th}}\) term, find ...

If mm times the mth{m^{th}} term of an A.P. is equal to nn times its nth{n^{th}} term, find the (m+n)th{(m + n)^{th}} term of the A.P.
A. 1 B. 0 C. - 1 D. 2  {\text{A}}{\text{. 1}} \\\ {\text{B}}{\text{. 0}} \\\ {\text{C}}{\text{. - 1}} \\\ {\text{D}}{\text{. 2}} \\\

Explanation

Solution

Hint:- nth{n^{th}}term of an A.P. is given by a+(n−1)da + (n - 1)d where d is the common difference between two consecutive terms. Using this formula write the terms given in the question therefore equating and rearranging those terms we will get the (m+n)th{(m + n)^{th}} term.

Complete step-by-step answer:
In this question
nth{n^{th}}term of given A.P. =a+(n−1)da + (n - 1)d eq1.
mth{m^{th}}term of given A.P. =a+(m−1)da + (m - 1)d eq2.
Now it is given that mm times the mth{m^{th}} term of an A.P. is equal to nn times its nth{n^{th}} term
Thus from eq1. , eq2.
⇒\Rightarrow ma+(m−1)d=na+(n−1)dm\\{ a + (m - 1)d\\} = n\\{ a + (n - 1)d\\}
On solving above equation
⇒ma+(m−1)md=na+(n−1)nd ⇒(m−n)a+(m2−m)d−(n2−n)d=0 ⇒(m−n)a+(m2−n2)d−(m−n)d=0   \Rightarrow ma + (m - 1)md = na + (n - 1)nd \\\ \Rightarrow (m - n)a + ({m^2}^{^{}} - m)d - ({n^2} - n)d = 0 \\\ \Rightarrow (m - n)a + ({m^2} - {n^2})d - (m - n)d = 0 \\\ \\\
On taking common from the above equation we get
⇒(m−n)a+(m+n)d−d=0 ⇒(m−n)a+(m+n−1)d=0   \Rightarrow (m - n)\\{ a + (m + n)d - d\\} = 0 \\\ \Rightarrow (m - n)\\{ a + (m + n - 1)d\\} = 0 \\\ \\\
Since m≠nm \ne n because mth{m^{th}} and nth{n^{th}} both terms are different
⇒a+(m+n−1)d=0\Rightarrow a + (m + n - 1)d = 0 eq 3.
Above equation represents the (m+n)th{(m + n)^{th}} term of the given A.P.
Hence option B is correct.

Note:- Whenever you get this type of question the key concept of solving is you have to write the given terms of A.P. in standard form like a+(n−1)da + (n - 1)d and then compare them. And get the resultant equation after solving the formed equation.