Question

the first term and common difference of an AP are respectively a and d . in another AP , the first term and the common difference are respectively a' and d'. if Sm=Sm' then prove that m=1+2(a-a')/d'-d (here Sm and Sm' are the respective sums of the two series upto m terms in each case).

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

We get that Sm and Sm' are same

Therefore,

m/2 [ 2a + (m-1)d ]  = m/2 [ 2a' + (m-1)d' ]

Therefore,

2a + md + d =  2a' + md' + d'

md - md' = 2a' - 2a + d' - d

m ( d - d') = 2(a' - a) + d' - d

m = [ 2(a' - a) + d' - d ]/ (d -d')

=  [2(a' - a)/ (d -d')]  +  [ -(d - d')/ (d - d')]

=  [2(a - a')/ (d' -d)] - 1

I hope this helps