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Question

The following is known about an AP:

Tm=1n,Tn=1m

Find the value of Smn. [4 MARKS]

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Solution

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+(m1)d=1n

Tn=a+(n1)d=1m

TmTn=(mn)d=1n1m

(mn)d=mnmn

d=1mn

Using this value of d in the any of the two initial relations, we obtain a=1mn.

Smn=(mn2)(2a+(mn1)d)

=(mn2){2mn+(mn1)(1mn)}

=mn+12

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