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.

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P

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?

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.

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 10^th 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., a_n is n^th term which is given by,

a_n = 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