Question

If the mth term of an A. P is 1/n and nth term is 1/m,then show that it's (mn)th term is 1

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BEST ANSWER WILL BE MARKED AS BRAINLIEST!

Answer 1

am = 1/n , an = 1/m

a+ (m-1)d = 1/n , a+ (n-1)d = 1/m

now, 1/n - 1/m = a+ (m-1)d -{a+ (n-1)d}

  (m-n)/mn = md-d - nd +d

  (m-n)/mn = (m-n)d

  d = 1 / (mn)

1 / n = a + (m-1)d

 1 / n   = a + (m-1) / mn

1 /n  = {amn + m-1} / mn

m = amn+m-1

a = 1 / mn

amn = a+ (mn-1)d

  = 1/mn + (mn-1) / mn

  = {1 + mn -1} / mn

  = mn / mn

 amn= 1

Answer 2

Answer:

mn th term of the AP is 1.

Step-by-step explanation:

It is given that the mth term of the AP is 1 / n and nth term of the AP is 1 / m.

We know( formula ),

xth term = a + ( x - 1 )d , where xth term is the last ( required ) term, a is the first term, x is the value of xth term and d is the common difference between the APs .

So,

mth term = a + ( m - 1 )d

= >  1  / n = a + dm - d      ...( i )  

       

Also,

nth term = a + ( n - 1 )d

= >  1 / m = a + dn - d       ...( ii )

Now, subtracting ( i ) from ( ii ),

= >  1 / m - 1 / n = a + dn - d - [ a + dm - d ]

= >  ( n - m ) / mn = a + dn - d - a - dm + d

= >  ( n - m ) / mn = dn - dm

= >  ( n - m ) / mn = d( n - m )

= >  1 / mn = d

Substituting the value of d in ( i ) ,

= >  1 / m = a + n( 1 / mn ) - 1 / mn

= >  1 / m = a + 1 / m - 1 / mn

= >  1 / mn = 1 / a

Now,

On the basis of the identity given above,

mn th term = a + ( mn - 1 )d

          ⇒ 1 / mn + ( mn - 1 )1 / mn

          ⇒ 1 / mn[ 1 + mn - 1 ]

          ⇒ 1 / mn [ mn ]

          ⇒ 1

Therefore the mn th term of the AP is 1.

Proved.