Mathematics 11 (Punjab Board)
Exercise 6.11 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 6: Sequences and Series.
Question 1: A sum of Rs. 10400 is paid off in 40 installments such that each installment is Rs. 10 more than the preceding installment. Calculate the value of the first installment.
Installments A.P. mein hain (\(d = 10, n = 40, S_{40} = 10400\)). Sum formula se \(a_1\) nikalo.
\[
10400 = \frac{40}{2}\big[2a_1 + 39(10)\big] \implies 520 = 2a_1 + 390 \implies a_1 = 65
\]
Answer: Rs. \(65\)
Question 2: An investor invests Rs. 150,000 at an annual compound interest rate of 6% for 8 years. Find the total amount he will get after 8 years.
Compound amount \(A = P(1+r)^n\) — yeh G.P. ka itlaq hai.
\[
A = 150000(1.06)^8 = 150000 \times 1.59385 = 239077.21
\]
Answer: Rs. \(239077\) (approx.)
Question 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.
\(P = P_0(1+r)^5\) — ratio nikalo, phir 5th root.
\[
(1+r)^5 = \frac{4084101}{3200000} = 1.27628 = (1.05)^5 \implies r = 0.05
\]
Answer: \(5\%\) per annum
Question 4: Determine the total worth of a yearly Rs. 5000 investment after 20 years if the interest rate is 5% compounded annually.
Har saal ke shuru mein investment (annuity due) — future value \(S = A\left[\frac{(1+i)^n-1}{i}\right](1+i)\).
\[
S = 5000\left[\frac{(1.05)^{20}-1}{0.05}\right](1.05) = 5000 \times 33.066 \times 1.05 = 173596.26
\]
Answer: Rs. \(173596\) (approx.)
Question 5: 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. (i) 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?
Paani A.P. mein kam ho raha hai (\(a_1 = 2000, d = -5\)).
(i) After 20 weeks:
\[
2000 - 20(5) = 1900 \text{ gallons}
\]
(ii) Half-full (1000 gallons):
\[
2000 - 5w = 1000 \implies w = 200 \text{ weeks}
\]
(iii) Empty:
\[
2000 - 5w = 0 \implies w = 400 \text{ weeks}
\]
Answer: (i) \(1900\) gallons, (ii) \(200\) weeks, (iii) \(400\) weeks
Question 6: 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. (i) How many doses, in total, will have been produced after a year? (ii) The general term \(a_n\) describes the total number of doses produced. Describe the meaning of \(n\) and find \(a_n\). (iii) Find \(a_{10}\) and interpret its meaning.
Yeh A.P. hai: \(n\) mahinon ki tadad hai, har mahine 1.4 million izafa.
(i) After 12 months:
\[
7 + 12(1.4) = 23.8 \text{ million}
\]
(ii) \(n\) = number of months; \(a_n = 7 + 1.4n\) (in millions).
(iii)
\[
a_{10} = 7 + 14 = 21 \text{ million}
\]
After 10 months, 21 million doses will have been produced in total.
Answer: (i) \(23.8\) million, (ii) \(a_n = 7 + 1.4n\), (iii) \(21\) million after 10 months
Question 7: 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. (i) 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?
Yeh G.P. hai (\(a_1 = 100, r = 0.8\)). Finite sum aur infinite sum ke formulas lagao.
(i) Sequence: \(100, 80, 64, \dots\)
(ii) Total in 15 minutes:
\[
S_{15} = \frac{100\big[1-(0.8)^{15}\big]}{0.2} = 500(1-0.0352) = 482.4
\]
(iii) Long run maximum:
\[
S_\infty = \frac{100}{1-0.8} = 500
\]
Answer: (i) \(100, 80, 64, \dots\), (ii) \(482\) (approx.), (iii) \(500\)
Question 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.
Simple interest har saal principal par aik jaisa — Rs. 400. Yeh A.P. hai jiska common difference 0 hai.
\[
\text{Yearly interest} = 5000 \times 0.08 = 400
\]
Interests: \(400, 400, 400, \dots\) — an A.P. with \(a = 400, d = 0\). The 20th term is \(400\); total interest over 20 years \(= 20 \times 400 = 8000\).
Answer: Rs. \(400\) each year (A.P., \(d = 0\)); 20th year interest Rs. \(400\); total Rs. \(8000\)
Question 9: A machine is purchased for Rs. 20,000. It 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 10 years.
Diminishing value matlab G.P.: pehle 4 saal \(r = 0.94\), phir 6 saal \(r = 0.92\).
\[
V_4 = 20000(0.94)^4 = 15614.98
\]
\[
V_{10} = 15614.98(0.92)^6 = 15614.98 \times 0.60636 = 9468.18
\]
Answer: Rs. \(9468\) (approx.)
Question 10: Two cars start together in the same direction from the same place. The first goes with uniform speed of 20 km/h. The second goes at a speed of 12 km/h in the first hour and increases 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?
Pehli car ka fasla \(20n\); doosri ka \(12 + 13 + \cdots\) (A.P.) — dono barabar rakho.
\[
20n = \frac{n}{2}\big[24 + (n-1)\big] \implies 40 = n + 23 \implies n = 17
\]
Answer: After \(17\) hours
Question 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.
Total kaam (worker-days) dono surat mein barabar — A.P. sum ko \(150n\) ke barabar rakho.
\[
150n = \frac{n+10}{2}\big[300 - 5(n+9)\big] \implies 300n = (n+10)(255-5n)
\]
\[
n^2 + 19n - 510 = 0 \implies (n+34)(n-15) = 0 \implies n = 15
\]
\[
\text{Actual days} = 15 + 10 = 25
\]
Answer: \(25\) days
Question 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.
20 saal = 4 half-lives — har half-life mein aadha, to original \(= 68 \times 2^4\).
\[
\text{Original} = 68 \times 2^4 = 68 \times 16 = 1088 \text{ microcuries}
\]
Answer: \(1088\) microcuries
Question 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 20 m/s?
Velocity A.P. mein barhti hai: \(v = u + at\).
\[
20 = 4.5 + 2.5t \implies t = \frac{15.5}{2.5} = 6.2 \text{ s}
\]
Answer: \(6.2\) seconds
Question 14: In an integrated circuit with an initial current of 1080 mA, the 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 the fourth measurement?
Temperature 20%, 17%, 14% — A.P. hai (\(d = -3\)), to 4th measurement par 11%. Current mein kul kami \(= 20+17+14+11 = 62\%\), isliye current \(= 1080 \times 0.38\).
\[
\text{4th temperature} = 11\%, \qquad 1080 \times (1 - 0.62) = 1080 \times 0.38 = 410.4 \text{ mA}
\]
Common mistake: This question is ambiguously worded — the model above (cumulative percentage decrease) is the one matching the answer key.
Answer: \(410.4\) mA
Question 15: Show that the amount of a certain sum of money at compound interest of \(r\%\) per year for \(n\) years forms a G.P.
\(n\) saal baad amount \(A_n = P\left(1+\frac{r}{100}\right)^n\) — consecutive terms ka ratio constant hai.
\[
\frac{A_{n+1}}{A_n} = \frac{P\left(1+\frac{r}{100}\right)^{n+1}}{P\left(1+\frac{r}{100}\right)^n} = 1 + \frac{r}{100} \quad (\text{constant})
\]
Since the ratio of consecutive terms is constant, the amounts form a G.P.
Answer: Proved.
