Formula: 1 Mark
Application: 2 Marks
Answer: 1 Mark
Let a and d be the first term and the common difference of the AP. We have:
Tm=a+(m−1)d=1n
Tn=a+(n−1)d=1m
⇒Tm−Tn=(m−n)d=1n−1m
⇒(m−n)d=m−nmn
⇒d=1mn
Using this value of d in the any of the two initial relations, we obtain a=1mn.
Smn=(mn2)(2a+(mn−1)d)
=(mn2){2mn+(mn−1)(1mn)}
=mn+12