PRACTICE QUESTIONS ON SEQUENCES AND SERIES

Question 1 :

A man saved ₹16500 in ten years. In each year after the first he saved ₹100 more than he did in the preceding year. How much did he save in the first year?

Solution :

Sum of money in ten years  =  16500

Let "a" be the first term.

By writing the savings in each year as sequence, we get

a, a + 100, a + 200,.........

Sn  =  (n/2)[2a + (n - 1) d]

a = a, d = 100 and n = 10

S₁₀  =  (10/2)[2a + (10 - 1) 100]

16500 =  5[2a + 900]

16500/5  =  2a + 900

  3300 - 900  =  2a

a  =  2400/2

a  =  1200

Question 2 :

Find the G.P. in which the 2nd term is √6 and the 6^th term is 9 √6.

Solution :

t₂  =  √6

ar  =  √6  ----(1)

t₆  =  9√6

ar⁵  =  9√6 ----(2)

(2)/(1)

ar⁵/ar  =  9√6/√6

r⁴  =  9

r⁴  =  (√3)⁴

r = √3

By applying the value of r in (1), we get 

a√3  =  √6

a  =  √6/√3

a  =  √2

√2, √6, 3√2,............

Hence the required G.P is √2, √6, 3√2,............

Question 3 :

The value of a motor cycle depreciates at the rate of 15% per year. What will be the value of the motor cycle 3 year hence, which is now purchased for  45,000?

Solution :

Present value of motor cycle  =  45000

a  =  45000

Since it is depreciated at the rate of 15% after one year the value of the machine is 85% of the initial value.

That is the value of the machine at the end of the first year is

  =  45000 x (85/100)

r  =  45000 x (85/100)

After two years, the value of the machine is 85% of the value in the first year.

Value of the machine at the end of the 2nd year is

  =  45000 × (85/100)²

Continuing this way, the value of the machine depreciates in the following way as

45000, 45000 x (85/100), 45000 × (85/100)², ................

The value of motor cycle after 3 years :

t_n  =  ar^n-1

t₄  =  45000  x (85/100)^4-1

t₄  =  45000  x (85/100)³

t₄  =  27636

Hence the cost of motor cycle after 3 years is 27636

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