Question

in a flower exhibition conducted at ooty for 4 days the number of tickets sold on the first,second ,third and forth days 110010,75070,25720 and 30636 draw the tree diagram​

Answer 1

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:

Answer 2

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)