if sum of 40 A.Ms between two numbers is 120,find the sum 50 AMs between them.

Let A_{1}, A_{2}, A_{3}, ........ , A_{40} be 40 A.M's between two numbers '*a*' and '*b*'.

Then, *a*, A_{1}, A_{2}, A_{3}, ........ , A_{40}, *b* is an A.P. with common difference *d* =

Now,

But sum of 40 A.M.'s is 120 [given]

∴ 120 = 20 (*a* + *b*)

⇒ (*a* + *b*) = 6 ............ (1)

Again consider B_{1}, B_{2}, ........ , B_{50} be 50 A.M.'s between two numbers *a* and *b*.

Then, *a*, B_{1}, B_{2}, ........ , B_{50}, *b* will be in A.P. with common difference

Now,

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