Q3 of 167 Page 19

A man arranges to pay off a debt of ₹ 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 instalment.

Given: Total debt is Rs. 3600 and total number of installments are 40 and in 30 installments, he has paid two-third of the debt and dies leaving one-third of amount


Let first instalment be “a”


Let the common difference of the instalments be “d”


Here, S40 = 3600 and n = 40


Formula used:



where a is first term, d is common difference and n is number of terms in an A.P.


Therefore,





⇒ 2a+39d = 180 ………(1)


Since 30 installments are paid and one third of the debt unpaid



So,


S30 = 2400 and n = 30







⇒ 2a+29d = 160 ………(2)


Solving (1) and(2) by substitution method,


2a + 39d = 180


⇒ 2a = 180 – 39d ……(3)


Put value of 2a from (3) in (2):


2a + 29d = 160


⇒ 180 – 39d + 29d = 160


⇒ –10d = 160 – 180


⇒ –10d = –20



⇒ d = 2


Put this value of d in (3):


2a = 180 – 39d


⇒ 2a = 180 – 39(2)


⇒ 2a = 180 – 78


⇒ 2a = 102



⇒ a = 51


Hence, value of first installment = a = Rs. 51


More from this chapter

All 167 →
1

A man saved ₹ 16500/- in ten years . in each year after the first he saved ₹ 100/- more then he did in the preceding year. How much did he saved in the first year?

 2

A man saves ₹ 32 during the first year, ₹ 36 in the second year and in this way he increases his savings by ₹4 every year. Find in what time his saving will be ₹200.

 4

A manufacturer of the radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find

(і) the production in the first year


(іі) the total product in the 7 years and


(ііі) the product in the 10th year.


Solution |||


(і) the production in the first year


Answer:


Given: 600 and 700 radio sets units are produced in third and seventh year respectively


To find: the production in the first year i.e. a


⇒ a3 = 600 and a7 = 700


Formula used:


For an A.P., an is nth term which is given by,


an = a + (n – 1)d


where a is first term, d is common difference and n is number of terms in an A.P.


Therefore,


a3 = a + (3 – 1)d


⇒ 600 = a + 2d……………………(1)


a7 = a + (7 – 1)d


⇒ 700 = a + 6d


⇒ a = 700 – 6d………………………(2)


Now put this value of a in equation (1):


⇒ 600 = 700 – 6d + 2d


⇒ 600 – 700 = – 6d + 2d


⇒ –100 = –4d



⇒ d = 25


Put d = 25 in equation (2):


⇒ a = 700 – 6(25)


⇒ a = 700 – 150


⇒ a = 550


Production in the first year = a = 550


(іі) the total product in the 7 years

 5

There are 25 trees at equal distances of 5 meters in a line with a well, the distance of well from the nearest tree being 10 meters. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.