arithmetic progression word problems?

Question

A man arranges to pay off the debt of Rs 3600 by 40 annnual 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?

Answer 1

3600=40(a1+an)/2
a1+an=180

2a1+39d=180

3/3-1/3=2/3

3600*2/3= 1200*2=2400 paid

a1+a30=160
a1+a1+29d=160

2a1+29d=160

by subtracting two equations
10d=20
d=2

2a1+29*2=160
2a1=102

answer:

a1=51
this the first installment

Answer 2

Let the value of the first instalment be Rs x.
Let the common difference of the A.P. be Rs d.
Then, the instalments are Rsx, Rs(x+d), Rs(x+2d),........ & the 30th instalment is Rs(x+29d)
Now, after 30 instalments are paid, one-third of the debt is left unpaid.
So, amount paid upto the 30 instalments = two-third of the debt
= 2/3 * Rs 3600
= Rs 2400
Thus, x + (x+d) + (x+2d) + ..........+(x + 29d) = 2400
i.e., (30/2) * [2x + (30-1)d]=2400
i.e., 2x + 29d = 160 -----------------(1)

Now, the man had planned to pay the debt in 40 instalments.
So, x + (x+d) + (x+2d) + .............+ (x+39d) = 3600
i.e., (40/2) * [2x + (40-1)d] = 3600
i.e., 2x + 39d = 180 -----------------(2)

Subtracting (1) from (2), we get,
10d = 20
i.e., d = 2 ------------------(3)

Substituting (3) in (1), we get,

Substituting (3) in (1), we get,
2x + 29*2 = 160
i.e., 2x + 58 = 160
i.e., 2x = 102
i.e., x = 51

Thus, the value of the 1st instalment is Rs 51.

Answer 3

90