Question

Ques-5
pls give me and. fast

Answer 1

Let the first installment be 'a'
We know that the installments are in A.P., let the common difference be 'd'

=> the installments are: a, a+d, a+2d,...a+39d
{as the nth term of an A.P. is [a+(n-1)d]}

Sum of n terms of an A.P. = n/2[a + (n-1) d]
=> S40 = 40/2(a + a + 39d)
3600 = 20 (2a + 39d)
[given that total debt is 3600]
=> 2a + 39d = 3600/20 = 180 - (1)

It is also given that 30 installments are paid. Unpaid amount = 1/3 of Rs.3600
= 1/3×3600
= 1200

Total payment till 30 installments = 3600-1200
= 2400
Rs.2400 is the amount of 30 installments i.e.
S30 = 30/2(a + a + 29d) = 2400
=> 15(2a + 29d) = 2400
=> 2a + 29d = 2400/15
=> 2a + 29d = 160 - (2)