Question

Ap question for my friends​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

HLO frnd here is ur answer.

Let a be the 1st term and d be the common difference. Then,

tm=1/n

a+(m-1)d = 1/n ———eq1

tn=1/m

a+(n-1)d= 1/m———eq2

subtracting eq 2 from eq 1, we get

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

[a+(m-1)d-a-(n-1)d]=m-n/mn

[d(m-1-n+1)]=m-n/mn

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

d=1/mn.

putting d=1/mn in eq 1, we get

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

a+(m-1)*1/mn=1/n

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

amn+m-1/mn=1/n

amn+m-1/1=mn/n

amn+m-1=m

amn-1=m-m

amn-1=0

amn=1

a=1/mn.

Therefore,

sum of mn terms,

Smn=mn/2[2a+(mn-1)d]

=mn/2[2*1/mn+(mn-1)*1/mn]

=mn/2[2/mn+mn-1/mn]

=mn/1[2+mn-1/mn]

=mn/2[mn+1/mn]

=mn/2*mn+1/mn

=mn+1/2.

hope it's help u