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If the mth term of an A.P. is 1/n and the nth term is 1/m, show that the sum of mn terms is (mn + 1) - Mathematics

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Sum

If the mth term of an A.P. is 1/n and the nth term is 1/m, show that the sum of mn terms is (mn + 1)

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Solution

Let a be the first term and d be the common difference of the given A.P. Then,

`a_m=1/n=>a+(m-1)d=1/n`

`a_n=1/n=>a+(n-1)d=1/n`

Subtracting equation (ii) from equation (i), we get

`(m-n)d=\frac{1}{n}-\frac{1}{m}`

`\Rightarrow (m-n)d=\frac{m-n}{mn}\Rightarrow d=\frac{1}{mn}`

Putting d = 1/mn in equation (i), we get

`a+(m-1)\frac{1}{mn}=\frac{1}{n}`

`\Rightarrow a+\frac{1}{n}-\frac{1}{mn}=\frac{1}{n}\Rightarrowa=\frac{1}{mn}`

Now,

`S_(mn)=(mn)/2{2a+(mn1)xxd}`

`S_(mn)=(mn)/2[2/(mn)+(mn-1)xx1/(mn)]`

`S_(mn)=1/2(mn+1)`

Concept: Sum of First n Terms of an AP
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