A man arranges to pay off a debt of Rs. 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one - third of the debt unpaid, find the value of the first instalment.
Let the 40 annual instalments form an arithmetic series of common difference d and first instalment a.
Then, series so formed is
a + (a + d) + (a + 2d) + .... =3600
Or Sn=n2[2a+(n−1)d]
Or 3600=20[2a+39d]
2a + 39d = 180 ..... (i)
and sum of first 30 terms is 23 of 3600
= 2400
⇒2400=302[2a+(29)d]
Or 2a + 29d = 160 .... (ii)
From (i) and (ii)
a = 51
The first instalment paid by this man is Rs. 51.