Progression (Sequence and Series)Hard
Question
-A man arrange to pay off a loan of Rs.3600 by 40 annual instalments which are in AP. When 30 of the instalments are paid, he dies leaving one third of the loan amount unpaid. The value of the 8th instalment is :-
Options
A.Rs. 35
B.Rs. 50
C.Rs. 65
D.Rs. 80
Solution
Let the first instalment be a and common difference of AP be d.
Given, 3600 = sum of 40 terms
=
[2a + (40 - 1)d]
⇒ 2a + 39d = 180 ....(i)
After 30 instalments one-third of the loan is unpaid.
Now
= 1200 is unpaid and 2400 is paid
So, 2400 =
[2a + (30 - 1)d]
⇒ 2a + 29d = 160 .....(ii)
Solving eq. (i) & (ii) a = 51, d = 2
So 8th instalment = a + (8 - 1) d = 65
Given, 3600 = sum of 40 terms
=
[2a + (40 - 1)d]⇒ 2a + 39d = 180 ....(i)
After 30 instalments one-third of the loan is unpaid.
Now
= 1200 is unpaid and 2400 is paidSo, 2400 =
[2a + (30 - 1)d]⇒ 2a + 29d = 160 .....(ii)
Solving eq. (i) & (ii) a = 51, d = 2
So 8th instalment = a + (8 - 1) d = 65
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