If mth term of an A.P. is \; 1/n and nth term is \; 1/m, then sum of mn terms is -

Correct Answer is : 12(mn+1)\; 1/2(mn+1)21(mn+1)

Correct Answer is : 12(mn+1)\; 1/2(mn+1)21(mn+1)

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Question : 28 of 86

Marks: +1, -0

\; \frac{1}{n}

\; \frac{1}{m}

, then sum of mn terms is -

Solution:

Concept:

Use standard formulas for

m

th and

n

th terms of an AP, then find sum of

mn

terms.

Explanation:

Let first term

a

and common difference

d

m

th term:

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

n

th term:

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

Subtract:

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

Thus

d = \frac{1}{mn}

(assuming

m \neq n

From

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

, we get

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

a = d = \frac{1}{mn}

Sum of

mn

terms:

S_{mn} = \frac{mn}{2}[2a + (mn-1)d] = \frac{mn}{2}\left[2 \cdot \frac{1}{mn} + (mn-1)\cdot\frac{1}{mn}\right] = \frac{mn}{2} \cdot \frac{mn+1}{mn} = \frac{mn+1}{2}

Answer:

\frac{1}{2}(mn+1)

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