arthmetic progression

I think that the wording of the question leaves much to be desired, but here's my interpretation of it.

As a preamble, the sum to n terms of the arithmetic progression with first term a and common difference d is (n/2)(2a + (n-1)d).

I take it that the first payment is a, the second a+d, the third a+2d and so on.

The intention was that 3600 was paid off in 40 instalments, so

3600 = (40/2)(2a + (40-1)d) , that is, 180 = 2a + 39d.

Two thirds are paid off after 30 instalments, so

2400 = (30/2)(2a + (30-1)d), or, 160 = 2a + 29d.

Solve those and you get a=51 and d=2.

You question does not make a lot of sense.  Perhaps you could re-word it.

Deleted - See Bertie's answer below.

yes brother,

one quarter of third.(1/3)

Mmmm..I wondered about this one yesterday, too !!! 

One fourth of one third = (1/12) * 3600 = 300......so he has paid 3300 after 30 payments

So, letting a₁ be the first payment and d be the monthly payment after that (i.e., the common difference), we have

a₁ + 39d = 3600

a₁ + 29d = 3300   ...multiply the second equation by -1 and add it to the first one

10d = 300  

d=30

a₁ + 29(30) = 300

a₁ + 870 = 300

a₁ = -570

I still come up with a negative initial payment.......maybe I'm doing something wrong......can anybody else help out??

I looked at this one before chris but I really do not understand the question.

Can you write your interpretation out for me. What is R?  What is 3600?  I have no idea.

Well done Bertie.  I forgot to sum the terms!!

Thanks, Bertie  !!!

so what is the first instalment?

Rs 51