Question -
Answer -
Given:
40 annual instalmentswhich form an arithmetic series.
Let the firstinstalment be ‘a’
S40 =3600, n = 40
By using the formula,
Sn =n/2 [2a + (n – 1)d]
3600 = 40/2 [2a +(40-1)d]
3600 = 20 [2a + 39d]
3600/20 = 2a + 39d
2a + 39d – 180 = 0 …..(i)
Given:
Sum of first 30 termsis paid and one third of debt is unpaid.
So, paid amount = 2/3× 3600 = ₹ 2400
Sn =2400, n = 30
By using the formula,
Sn =n/2 [2a + (n – 1)d]
2400 = 30/2 [2a +(30-1)d]
2400 = 15 [2a + 29d]
2400/15 = 2a + 29d
2a + 29d -160 = 0 ….(ii)
Now, let us solveequation (i) and (ii) by substitution method, we get
2a + 39d = 180
2a = 180 – 39d … (iii)
Substitute the valueof 2a in equation (ii)
2a + 29d – 160 = 0
180 – 39d + 29d – 160= 0
20 – 10d = 0
10d = 20
d = 20/10
= 2
Substitute the valueof d in equation (iii)
2a = 180 – 39d
2a = 180 – 39(2)
2a = 180 – 78
2a = 102
a = 102/2
= 51
Hence, value of firstinstallment ‘a’ is ₹ 51