Question

A man arranges to pay a debt of Rs.3600 in 40 monthly installments
which are in an AP.When 30 installments are paid he dies leaving one third of the debt unpaid.Find Value The first installment.
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answer 1

Let the first term (instalment ) be ₹ a.

Let the common difference between successive instalment be ₹ d

In  case 1:



n = 40 and  S40 = ₹ 3600


Sn = n/2[2a+(n-1)d]


S40= 40/2[2a+(40-1)d]


3600 = 20[2a+39d]


3600/20= 2a+39d


180 = 2a+39d


2a+39d=180……………(1)


In case 2:


Debt unpaid= (⅓)×3600


Debt unpaid= ₹ 1200


Debt paid by man in 30 installments= Total debt paid- debt paid

Debt paid by man in 30 installments=3600 – 1200 = ₹ 2400

n = 30 &  S30 = 2400

Sn = n/2[2a+(n-1)d]


S30 = 30/2[ 2a+ (30-1)d]


2400= 15[2a+29d]


2400/15 = 2a+29d


160 = 2a+29d…………...(2)

Subtract eq (2) from eq (1)

(2a + 39d) – (2a + 29d) = 180 – 160


2a + 39d – 2a - 29d = 20

39d – 29d = 20

10 d = 20

d = 20/10


d= 2

Put the value of d in eq (1)

2a + 39d = 180

2a + 39 × 2 = 180

2a = 180 – 78

2a = 102


a= 102/2

a = 51

Hence, the first instalment paid by man = ₹ 51



Hope this will help you....