Question

If the ath term of an AP be 1/b and the bth term be 1/a then show that its (ab)th term is 1.

Answer 1

Step-by-step explanation:

Given that, mth term=1/n and nth term=1/m.

then ,let a and d be the first term and the common difference of the A.P.

so a+(m-1)d=1/n...........(1) and a+(n-1)d=1/m...........(2).

subtracting equation (1) by (2) we get,

md-d-nd+d=1/n-1/m

=>d(m-n)=m-n/mn

=>d=1/mn.

again if we put this value in equation (1) or (2) we get, a=1/mn.

then, let A be the mnth term of the AP

a+(mn-1)d=1/mn+1+(-1/mn)=1

hence proved.

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Answer 2

Answer:

1/b =A + (a-1)D .....eq(1)

1/a= A +(b-1)D ......eq(2)

subtracting eq1 and 2

1/b-1/a=(a-1-b+1)D

a-b/ab=(a-b)D

taking (a-b) common both side

1/ab=D .......eq(3)

using eq(3) in eq (1)

1/b = A +(1/b-1/ab)

on solving further we get

A= 1/ab .......eq(4)

regarding to question we have to find (ab)th term

(?)=A+(ab-1)D

substituting the eq(3) &eq(4) values of A and D

(?)= 1/ab + (ab-1)1/ab

(?)= (1+ab -1)/ab

(?)=ab/ab

(?)= 1

HENCE PROVED THE (ab)th TERM =1