Test Index

Sequence and Series

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Question : 32 of 86

Marks: +1, -0

If the mth term of A.P. is

\frac{1}{n}

and the nth term is

\frac{1}{m},

then its mnth term is:

1
$-1$
0
None

Solution:

Concept:

Use the formula for the

n

th term of an arithmetic progression:

T_n = a + (n-1)d

.

Explanation:

Let

a

be the first term and

d

the common difference.

Given

T_m = \frac{1}{n}

:

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

(1)

Given

T_n = \frac{1}{m}

:

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

(2)

Subtract (2) from (1):

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

.

Since

m \neq n

,

d = \frac{1}{mn}

.

Substitute in (1):

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

⟹

a + \frac{m-1}{mn} = \frac{m}{mn}

⟹

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

.

Now

T_{mn} = a + (mn-1)d = \frac{1}{mn} + (mn-1)\cdot\frac{1}{mn} = \frac{1 + mn - 1}{mn} = \frac{mn}{mn} = 1

.

Answer:

1

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