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Learn/ 11th Class Mathematics/ Unit 6 /Numericals

Unit 6: Sequences and Series — Exercise 6 11

11th Class Mathematics · Unit 6: Sequences and Series

1.A sum of Rs. 10400 is paid off in 40 instalments such that each instalment is Rs. 10 more than the preceding instalment. Calculate the value of the first instalment.
Given
Sum Sn = 10400, n = 40, d = 10
Formula
Arithmetic progression sum Sn = n2[2a + (n-1)d]
Working
Substituting values 10400 = 402[2a + (40-1) × 10]
10400 = 20[2a + 390]
520 = 2a + 390
2a = 130
Result
First instalment a = 65 Rs
2.An investor invests Rs. 150000 at an annual compound interest rate of 6% for 8 years. Find the total amount will he get after 8 years.
3.The population of a town is 4084101 at present and five years ago it was 3200000. Find its rate of increase if it increased geometrically.
Given
Population data a1 = 3200000, a6 = 4084101
Formula
Geometric progression an = a1 rn-1
Working
Setting up equation 4084101 = 3200000 × r5
r5 = 40841013200000 = 1.2762...
r = 1.05
Result
Rate of increase r = 1 + 5100 implies 5%
4.Determine the total worth of a yearly Rs. 5000 investment after 20 years if the interest rate is 5% compounded annually.
Given
Investment details a1 = 5000, r = 1.05, n = 20
Formula
Geometric series sum Sn = a1(rn - 1)r - 1
S20 = 5000 × 1.05[(1.05)20 - 1]1.05 - 1
= 5250[2.6533 - 1]0.05
= 5250 × 1.65330.05 = 5250 × 33.066
Result
Total amount after 20 years S20 ≈ 173596 Rs
5.(i) A water tank has a leakage. Each week, the tank loses 5 gallons of water due to the leakage. Initially, the tank is full and contains 2000 gallons. How many gallons are in the tank 20 weeks later? (ii) How many weeks until the tank is half-full? (iii) How many weeks until the tank is empty?
Given
(i) Tank parameters a1 = 2000, d = -5, n = 20
Formula
Arithmetic progression an+1 = a1 + nd
Working
After 20 weeks a21 = 2000 + 20 × (-5)
= 2000 - 100
Result
Water remaining = 1900 gallons
Given
(ii) Half-full condition an+1 = 1000
Formula
Arithmetic term an+1 = a1 + nd
Working
Setting up equation 1000 = 2000 + n(-5)
1000 = 2000 - 5n
5n = 1000
Result
Weeks to half-full n = 200 weeks
Given
(iii) Empty condition an+1 = 0
Formula
Arithmetic term an+1 = a1 + nd
Working
Setting up equation 0 = 2000 + n(-5)
0 = 2000 - 5n
5n = 2000
Result
Weeks to empty n = 400 weeks
6.(i) A drug company has manufactured 7 million doses of a vaccine to date. They promise additional production at a rate of 1.4 million doses/month over the next year. How many doses of the vaccine, in total, will have been produced after a year? (ii) The general term a_n describes the total number of doses of the vaccine produced. Describe the meaning of the variable n in the context of this problem. Find the general term a_n. (iii) Find the value of a_10 and interpret its meaning in words.
Given
(i) Initial and monthly production Initial = 7, Monthly = 1.4 million
After 1 month 7 + 1.4 = 8.4 million
After 2 months 8.4 + 1.4 = 9.8 million
After 12 months 7 + 12 × 1.4 = 7 + 16.8
Result
Total after 1 year = 23.8 million doses
Given
(ii) Interpretation of n n represents the number of months elapsed
Formula
General term for linear sequence an = a1 + (n-1)d
an = 7 + 1.4n (million doses after n months)
Result
General term an = 7 + 1.4n
Given
(iii) General term an = 7 + 1.4n
Working
Substituting n = 10 a10 = 7 + 1.4 × 10
= 7 + 14
Result
Production after 10 months a10 = 21 million doses
7.(i) At a toll booth, the number of vehicles passing through during the first minute is 100. Due to road congestion, each minute only 80% of the vehicles from the previous minute manage to pass. Represent the number of vehicles passing each minute as a sequence. (ii) Find the total number of vehicles that pass through in 15 minutes. (iii) What is the maximum number of vehicles that can pass in the long run (as time t → ∞)?
Given
(i) First minute a1 = 100
Each subsequent minute 80% of previous minute
Second minute a2 = 100 × 80100 = 80
Third minute a3 = 80 × 80100 = 64
Result
GP with ratio a1 = 100, r = 0.8; Sequence: 100, 80, 64, 51.2, ldots
Given
(ii) GP parameters a1 = 100, r = 0.8, n = 15
Formula
Geometric series sum Sn = a1(1 - rn)1 - r
Working
Substituting values S15 = 100[1 - (0.8)15]1 - 0.8
= 100[1 - 0.03518]0.2
= 500[1 - (0.8)15] = 500 × 0.96482
Result
Total vehicles in 15 minutes S15 = 482.4
Formula
(iii) Infinite GP sum S_infty = a11 - r
Working
With |r| < 1 S_infty = 1001 - 0.8
= 1000.2
Result
Maximum vehicles = 500 vehicles
8.A sum of Rs. 5000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an A.P.? If so, find the interest at the end of 20 years making use of this fact.
Given
Investment and rate P = 5000, r = 8%
Interest per year I = 5000 × 0.08 = 400 Rs
Sequence of interests 400, 400, 400, 400, ldots
Formula
AP with a = 400, d = 0 an = a + (n-1)d
Working
Total interest for 20 years S20 = 202[2 × 400 + (20-1) × 0]
= 10 × 800
Result
Total interest after 20 years = 8000 Rs
9.A machine is purchased for Rs.20,000. Depreciates at 6% per annum for the first four years and after that 8% per annum for the next six years. Depreciation being calculated on diminishing value. Find the value of the machine after a period of 10 years.
Given
Initial value a1 = 20000
First 4 years r1 = 1 - 0.06 = 0.94
Formula
After 4 years a5 = a1 × r14
a5 = 20000 × (0.94)4 = 15614.98
Given
Next 6 years b1 = 15614.98, r2 = 0.92
Formula
After 6 more years b7 = b1 × r26
= 15614.98 × (0.92)6 = 9468.22
Result
Machine value after 10 years = Rs. 9468.22
10.Two cars start together in the same direction from the same place. The first goes with uniform speed of 20km/h. The second goes at a speed of 12km/h in the first hour and increase the speed by 1 km/h each succeeding hour. After how many hours will the second car overtake the first car if both cars go non-stop?
Given
First car Uniform speed = 20 km/h
Second car hourly speeds 12, 13, 14, ldots km/h
Formula
Distance covered by first car D1 = 20n
Distance covered by second car D2 = Sn = n2[2 × 12 + (n-1) × 1]
= n2[24 + n - 1] = n2[23 + n]
Working
For overtaking n2[23 + n] = 20n
23n + n2 = 40n
n2 - 17n = 0
n(n - 17) = 0
Result
Time for overtaking n = 17 hours
11.150 workers were engaged to finish a piece of work in a certain number of days. Five workers dropped the second day, five more workers dropped the third day and so on. It takes 10 more days to finish the work now. Find the number of days in which the work was completed.
Given
Initial workers 150
Workers sequence 150, 145, 140, 135, ldots
This is an AP a1 = 150, d = -5
Original work 150n
Formula
Work completed in (n+10) days Sn+10 = n+102[2 × 150 + (n+9)(-5)]
= n+102[300 - 5n - 45]
= n+102[255 - 5n]
Working
Setting equal 150n = n+102[255 - 5n]
300n = (n+10)(255 - 5n)
n2 + 19n - 510 = 0
(n + 34)(n - 15) = 0
n = 15 (since n > 0)
Result
Total days with additional 10 days n + 10 = 25 days
12.A radioactive product has a half-life of 5 years. If the radioactivity level is 68 microcuries after 20 years. Determine the original level of radioactivity.
Given
Half-life 5 years
Level after 20 years as = 68 microcuries
Formula
Geometric progression an = a1 rn-1
After each 5-year period r = 12
Number of half-lives in 20 years 20 div 5 = 4
Working
After 4 half-lives a5 = a1 × (12)4
68 = a1 × 116
Result
Original radioactivity a1 = 68 × 16 = 1088 microcuries
13.An object moving in a line is given an initial velocity of 4.5 m/s and a constant acceleration of 2.5 m/s². How long will it take the object to reach a velocity of 20m/s?
14.In an integrated circuit with an initial current of 1080 mA, temperature in the components decreases from 20% to 17% to 14%. Assuming that each temperature decrease is caused by a decrease in the initial current, what is the value of the current at fourth measurement?
Given
Initial current I1 = 1080 mA
Temperature decreases 20%, 17%, 14%, 11%
These form an AP a1 = 20, d = -3
Total decrease 20 + 17 + 14 + 11 = 62%
Working
Current at 4th measurement I4 = 1080 × (1 - 0.62)
= 1080 × 0.38
Result
Current at fourth measurement = 410.4 mA
15.Show that the amount of a certain sum of money at compound interest of r% per year for n years form a G.P.
Given
Principal P
Rate of interest r% per year
Interest for first year I1 = Pr100
Amount for first year A1 = P + Pr100 = P(1 + r100)
Amount for second year A2 = P(1 + r100)2
Amount for third year A3 = P(1 + r100)3
Amount for n years An = P(1 + r100)n
Result
Sequence of amounts P(1 + r100), P(1 + r100)2, ldots, P(1 + r100)n is a G.P with first term P(1 + r100) and common ratio (1 + r100)