If the mth term of an AP is 1/n and nth term is 1/m then show that sum of mn term is 1/2(mn +1)
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Answer:
Let mth term of AP be ‘Am’ and nth term of AP be ‘An’
Therefore, Am = a+ (m-1)d=1/n ….(i)
An = a+(n-1)d=1/m ….(ii)
Subtracting equation (ii) from (i)
d[(m-1)-(n-1)] = 1/n-1/m,
d(m-n) = (m-n)/mn,
d = 1/mn ….(iii)
Substituting equation (iii) in (i)
a+(m-1)/mn = 1/n,
a = 1/n[1-(m-1)/m],
a = 1/mn ….(iv)
Now Amn i.e the mnth term of AP = a+(mn-1)d,
Substitute equation (iii) and (iv) in Amn,
1/mn+(mn-1)/mn = 1/mn[1+(mn-1)] = mn/mn = 1,
then the mn term = 1
sum of mn term :-
Amn = mn/2 ( 2/mn + (mn-1)1/mn)
= 1 + (mn)/2 - 1/2
= mn /2 + 1/2
= 1/2 (mn + 1)
Step-by-step explanation:
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Let mth term of AP be ‘Am’ and nth term of AP be ‘An’
Therefore, Am = a+ (m-1)d=1/n ….(i)
An = a+(n-1)d=1/m ….(ii)
Subtracting equation (ii) from (i)
d[(m-1)-(n-1)] = 1/n-1/m,
d(m-n) = (m-n)/mn,
d = 1/mn ….(iii)
Substituting equation (iii) in (i)
a+(m-1)/mn = 1/n,
a = 1/n[1-(m-1)/m],
a = 1/mn ….(iv)
Now Amn i.e the mnth term of AP = a+(mn-1)d,
Substitute equation (iii) and (iv) in Amn,
1/mn+(mn-1)/mn = 1/mn[1+(mn-1)] = mn/mn = 1,
then the mn term = 1
sum of mn term :-
Amn = mn/2 ( 2/mn + (mn-1)1/mn)
= 1 + (mn)/2 - 1/2
= mn /2 + 1/2
= 1/2 (mn + 1)
Step-by-step explanation: