Sn = n/2{2a+(n-1)d}

Where n= no. of terms of the AP

a= first term

d = difference b/w two consecutive terms

hence for n=40

20{2a+39d}= 3600

2a+39d=180 —(1)

Similarly when 30 installments have been paid the remaining amount is 1/3rd so paid amount is 2/3rd of 3600 which is 2/3*3600= 2400 hence for n=30

15{2a+29d}=2400

2a+29d=160 –(2)

Solving equation 1 & 2

a=51

d=2

now nth term of a AP is

Tn=a+(n-1)d

Since the old man paid installment i.e. first term of AP hence the old man paid “51” as first installment.

The heirs paid the 31st installment i.e. 31st term of AP

Which is

T31= 51+30*2=111

Hence the heirs paid “111” as first installment.

Value of first installment paid by

(i)Mr. A = 51

(ii)the heirs = 111