Given: –
Total debt = Rs.36000
A man pays this debt in 40 annual instalments that forms an A.P.
After annual instalments, that man dies leaving one – third of the debt unpaid.
So, Within 30 instalments he pays two – thirds of his debt.
Sum of n terms in an Arithmetic Progression =
n2[2a+(n–1)d]He has to pay 36000 in 40 annual instalments,
36000=402[2×a+(40–1)×d]Where,
a = amount paid in the first instalment,
d = difference between two Consecutive instalments.
He paid two – a third of the debt in 30 instalments,
23(36000)=302[2a+(30–1)d]From equations (1) & (2) we get,
a = 510 & d = 20
∴The value of the first instalment is Rs.510.