Welcome to QPD (Questions Per Day) series. Today’s question is taken from Mathematics, chapter - “Sequence & Series”.
If \(a_m=\frac{1}{n}\) and \(a_n=\frac{1}{m}\) for an AP, find \(a_{mn}\).
Solution:
\(a_m=\frac{1}{n}\)
\(\implies a_m = a+(m-1)d=\frac{1}{n}\) ...equation (i)
\(a_n=\frac{1}{m}\)
\(\implies a_n = a+(n-1)d=\frac{1}{m}\) ...equation (ii)
Now eliminate the \(a\) by subtracting equation (ii) in (i) or vice-versa.
\((n-1)d-(m-1)d=\frac{1}{m}-\frac{1}{n}\)
\(\implies (n-m)d = \frac{(n-m)}{mn}\)
\(\implies \boxed{d = \frac{1}{mn}}\)
Now, we will find \(a\) by putting \(d\) in equation (i) or (ii)
Let’s put in equation (i)
\(a+(m-1)\frac{1}{mn} =\frac{1}{n}\)
\(\implies a+\frac{m-1}{mn}=\frac{1}{n}\)
\(\implies a = \frac{1}{n}-\frac{m-1}{mn}\)
\(\implies a = \frac{1}{n}-\frac{1}{n}+\frac{1}{mn}\)
\(\implies \boxed{a=\frac{1}{mn}}\)
Now we’ve the values for \(a\) and \(d\), we will find \(a_{mn}\)
\(a_{mn} = a+(mn-1)d\)
Put the values of \(a\) and \(d\)
\(\implies a_{mn} = \frac{1}{mn}+\frac{mn-1}{mn}\)
\(\implies a_{mn} = \frac{1}{mn}+\frac{mn}{mn}-\frac{1}{mn}\)
\(\implies \boxed{a_{mn}=1}\)
Comments
Post a Comment