Class 11: Arithmetic Progressions

$\displaystyle \textbf{Question 1: }\text{A man saved Rs. }16500\text{ in ten years. In each year after the first, he saved}$
$\displaystyle \text{Rs. }100\text{ more than in the preceding year. How much did he save in the first year?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the amount saved by the man in the first year be Rs. }x.$
$\displaystyle \therefore x,(x+100),(x+200),\ldots,(x+900)\text{ are his yearly savings.}$
$\displaystyle x+(x+100)+(x+200)+\cdots+(x+900)=16500$
$\displaystyle 10x+(100+200+\cdots+900)=16500$
$\displaystyle 10x+\frac{9}{2}\left[2(100)+(9-1)(100)\right]=16500$
$\displaystyle 10x+4500=16500$
$\displaystyle 10x=12000$
$\displaystyle \therefore x=1200$
$\displaystyle \therefore \text{The man saved Rs. }1200\text{ in the first year.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{A man saves Rs. }32\text{ during the first year, Rs. }36\text{ in the second year,}$
$\displaystyle \text{and increases his savings by Rs. }4\text{ every year. Find the time in which his total}$
$\displaystyle \text{savings will be Rs. }200.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The yearly savings form the A.P. }32,36,40,\ldots$
$\displaystyle a=32,\qquad d=36-32=4,\qquad S_n=200$
$\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]$
$\displaystyle 200=\frac{n}{2}\left[2(32)+(n-1)(4)\right]$
$\displaystyle 200=\frac{n}{2}(64+4n-4)$
$\displaystyle 200=n(2n+30)$
$\displaystyle 2n^2+30n-200=0$
$\displaystyle n^2+15n-100=0$
$\displaystyle (n-5)(n+20)=0$
$\displaystyle n=5\quad\text{or}\quad n=-20$
$\displaystyle \text{Since the number of years cannot be negative, }n=5.$
$\displaystyle \therefore \text{His total savings will be Rs. }200\text{ in }5\text{ years.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{A man arranges to pay off a debt of Rs. }3600\text{ in }40\text{ annual instalments}$
$\displaystyle \text{which form an A.P. After paying }30\text{ instalments, he dies, leaving one-third of the}$
$\displaystyle \text{debt unpaid. Find the value of the first instalment.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the first instalment be Rs. }a\text{ and the common difference be Rs. }d.$
$\displaystyle \text{The sum of all }40\text{ instalments is Rs. }3600.$
$\displaystyle S_{40}=\frac{40}{2}\left[2a+(40-1)d\right]$
$\displaystyle 3600=20(2a+39d)$
$\displaystyle 2a+39d=180\qquad\ldots\text{(i)}$
$\displaystyle \text{One-third of the debt remained unpaid.}$
$\displaystyle \therefore \text{Amount paid in the first }30\text{ instalments}=3600-\frac{1}{3}(3600)$
$\displaystyle =3600-1200=2400\text{ Rs.}$
$\displaystyle S_{30}=\frac{30}{2}\left[2a+(30-1)d\right]$
$\displaystyle 2400=15(2a+29d)$
$\displaystyle 2a+29d=160\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting (ii) from (i),}$
$\displaystyle 10d=20$
$\displaystyle d=2$
$\displaystyle \text{Substituting }d=2\text{ in (i),}$
$\displaystyle 2a+39(2)=180$
$\displaystyle 2a=102$
$\displaystyle \therefore a=51$
$\displaystyle \therefore \text{The value of the first instalment is Rs. }51.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{A manufacturer of radio sets produced }600\text{ units in the third year and }700$
$\displaystyle \text{units in the seventh year. Assuming that production increases uniformly by a fixed number}$
$\displaystyle \text{every year, find: (i) the production in the first year (ii) the total production in }7\text{ years}$
$\displaystyle \text{and (iii) the production in the }10^{\text{th}}\text{ year.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a_3=600\text{ and }a_7=700$
$\displaystyle a_n=a+(n-1)d$
$\displaystyle 600=a+(3-1)d$
$\displaystyle a+2d=600\qquad\ldots\text{(i)}$
$\displaystyle 700=a+(7-1)d$
$\displaystyle a+6d=700\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting (i) from (ii),}$
$\displaystyle 4d=100$
$\displaystyle \therefore d=25$
$\displaystyle \text{Substituting }d=25\text{ in (i),}$
$\displaystyle a+2(25)=600$
$\displaystyle \therefore a=550$
$\displaystyle \text{i) Production in the first year}=550\text{ units}$
$\displaystyle \text{ii) }S_7=\frac{7}{2}\left[2a+(7-1)d\right]$
$\displaystyle =\frac{7}{2}\left[2(550)+6(25)\right]$
$\displaystyle =\frac{7}{2}(1100+150)$
$\displaystyle =\frac{7}{2}\times1250$
$\displaystyle \therefore S_7=4375\text{ units}$
$\displaystyle \text{iii) }a_{10}=a+(10-1)d$
$\displaystyle =550+9(25)$
$\displaystyle =550+225$
$\displaystyle \therefore a_{10}=775\text{ units}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{There are }25\text{ trees in a line at equal distances of }5\text{ metres. A well is}$
$\displaystyle 10\text{ metres from the nearest tree. A gardener waters each tree separately, starting from}$
$\displaystyle \text{the well and returning to it after watering each tree. Find the total distance covered.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The distance of the first tree from the well is }10\text{ m.}$
$\displaystyle \therefore \text{Distance covered to water the first tree and return}=2(10)=20\text{ m.}$
$\displaystyle \text{The round-trip distances form the A.P. }20,30,40,\ldots$
$\displaystyle a=20,\qquad d=30-20=10,\qquad n=25$
$\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]$
$\displaystyle S_{25}=\frac{25}{2}\left[2(20)+(25-1)(10)\right]$
$\displaystyle =\frac{25}{2}(40+240)$
$\displaystyle =\frac{25}{2}\times280$
$\displaystyle \therefore S_{25}=3500\text{ m}$
$\displaystyle \therefore \text{The total distance covered by the gardener is }3500\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{A man is employed to count Rs. }10710.\text{ He counts at the rate of Rs. }180$
$\displaystyle \text{per minute for half an hour. After this, he counts Rs. }3\text{ less every minute than in the}$
$\displaystyle \text{preceding minute. Find the time taken by him to count the entire amount.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Amount counted in the first }30\text{ minutes}=180\times30$
$\displaystyle =5400\text{ Rs.}$
$\displaystyle \text{Amount remaining}=10710-5400$
$\displaystyle =5310\text{ Rs.}$
$\displaystyle \text{Thereafter, the amounts counted per minute form the A.P. }177,174,171,\ldots$
$\displaystyle a=177,\qquad d=174-177=-3,\qquad S_n=5310$
$\displaystyle \text{Let the time taken to count the remaining amount be }n\text{ minutes.}$
$\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]$
$\displaystyle 5310=\frac{n}{2}\left[2(177)+(n-1)(-3)\right]$
$\displaystyle 5310=\frac{n}{2}(354-3n+3)$
$\displaystyle 10620=n(357-3n)$
$\displaystyle 3n^2-357n+10620=0$
$\displaystyle n^2-119n+3540=0$
$\displaystyle (n-59)(n-60)=0$
$\displaystyle n=59\quad\text{or}\quad n=60$
$\displaystyle \text{For }n=60,\text{ the amount counted in the }60^{\text{th}}\text{ minute is}$
$\displaystyle a_{60}=177+(60-1)(-3)=0.$
$\displaystyle \text{Thus, the entire remaining amount is already counted in }59\text{ minutes.}$
$\displaystyle \therefore \text{Total time taken}=30+59$
$\displaystyle =89\text{ minutes}$
$\displaystyle \therefore \text{The time taken to count the entire amount is }89\text{ minutes.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{A piece of equipment cost a factory Rs. }600000.\text{ It depreciates by }15\%$
$\displaystyle \text{in the first year, }13.5\%\text{ in the second year, }12\%\text{ in the third year, and so on.}$
$\displaystyle \text{Find its value at the end of }10\text{ years, if all percentages apply to the original cost.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The yearly depreciation percentages form the A.P. }15,13.5,12,\ldots$
$\displaystyle a=15,\qquad d=13.5-15=-1.5,\qquad n=10$
$\displaystyle \text{Total depreciation percentage}=S_{10}$
$\displaystyle S_{10}=\frac{10}{2}\left[2(15)+(10-1)(-1.5)\right]$
$\displaystyle =5(30-13.5)$
$\displaystyle =5(16.5)$
$\displaystyle =82.5\%$
$\displaystyle \text{Total depreciation}=\frac{82.5}{100}\times600000$
$\displaystyle =495000\text{ Rs.}$
$\displaystyle \text{Value after }10\text{ years}=600000-495000$
$\displaystyle \therefore \text{The value of the equipment after }10\text{ years is Rs. }105000.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{A farmer buys a used tractor for Rs. }12000.\text{ He pays Rs. }6000\text{ in cash}$
$\displaystyle \text{and agrees to pay the balance in annual instalments of Rs. }500\text{ plus }12\%\text{ interest}$
$\displaystyle \text{on the unpaid amount. Find the total amount paid for the tractor.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Cost of the tractor}=12000\text{ Rs.}$
$\displaystyle \text{Cash payment}=6000\text{ Rs.}$
$\displaystyle \text{Unpaid principal}=12000-6000=6000\text{ Rs.}$
$\displaystyle \text{The principal repaid annually is Rs. }500.$
$\displaystyle \therefore \text{Number of annual instalments}=\frac{6000}{500}=12$
$\displaystyle \text{The interest payments are}$
$\displaystyle 6000\left(\frac{12}{100}\right),5500\left(\frac{12}{100}\right),5000\left(\frac{12}{100}\right),\ldots$
$\displaystyle \text{i.e. }720,660,600,\ldots,60.$
$\displaystyle \text{These interest payments form an A.P.}$
$\displaystyle a=720,\qquad d=660-720=-60,\qquad n=12$
$\displaystyle S_{12}=\frac{12}{2}\left[2(720)+(12-1)(-60)\right]$
$\displaystyle =6(1440-660)$
$\displaystyle =6(780)$
$\displaystyle =4680\text{ Rs.}$
$\displaystyle \text{Total amount paid}=\text{cost of the tractor}+\text{total interest}$
$\displaystyle =12000+4680$
$\displaystyle \therefore \text{The tractor cost him Rs. }16680.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Shamshad Ati buys a scooter for Rs. }22000.\text{ He pays Rs. }4000\text{ in cash}$
$\displaystyle \text{and agrees to pay the balance in annual instalments of Rs. }1000\text{ plus }10\%\text{ interest}$
$\displaystyle \text{on the unpaid amount. Find the total amount paid for the scooter.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Cost of the scooter}=22000\text{ Rs.}$
$\displaystyle \text{Cash payment}=4000\text{ Rs.}$
$\displaystyle \text{Unpaid principal}=22000-4000=18000\text{ Rs.}$
$\displaystyle \text{The principal repaid annually is Rs. }1000.$
$\displaystyle \therefore \text{Number of annual instalments}=\frac{18000}{1000}=18$
$\displaystyle \text{The interest payments are}$
$\displaystyle 18000\left(\frac{10}{100}\right),17000\left(\frac{10}{100}\right),16000\left(\frac{10}{100}\right),\ldots$
$\displaystyle \text{i.e. }1800,1700,1600,\ldots,100.$
$\displaystyle \text{These interest payments form an A.P.}$
$\displaystyle a=1800,\qquad d=1700-1800=-100,\qquad n=18$
$\displaystyle S_{18}=\frac{18}{2}\left[2(1800)+(18-1)(-100)\right]$
$\displaystyle =9(3600-1700)$
$\displaystyle =9(1900)$
$\displaystyle =17100\text{ Rs.}$
$\displaystyle \text{Total amount paid}=\text{cost of the scooter}+\text{total interest}$
$\displaystyle =22000+17100$
$\displaystyle \therefore \text{The scooter cost him Rs. }39100.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{The income of a person is Rs. }300000\text{ in the first year, and it increases}$
$\displaystyle \text{by Rs. }10000\text{ every year for the next }19\text{ years. Find the total income received}$
$\displaystyle \text{in }20\text{ years.}$
$\displaystyle \text{Answer:}$
$\displaystyle a=300000,\qquad d=10000,\qquad n=20$
$\displaystyle S_{20}=\frac{20}{2}\left[2(300000)+(20-1)(10000)\right]$
$\displaystyle =10(600000+190000)$
$\displaystyle =10(790000)$
$\displaystyle =7900000\text{ Rs.}$
$\displaystyle \therefore \text{The total income received in }20\text{ years is Rs. }7900000.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{A man starts repaying a loan with a first instalment of Rs. }100.\text{ If he}$
$\displaystyle \text{increases the instalment by Rs. }5\text{ every month, find the amount of the }30^{\text{th}}$
$\displaystyle \text{instalment.}$
$\displaystyle \text{Answer:}$
$\displaystyle a=100,\qquad d=5,\qquad n=30$
$\displaystyle a_{30}=a+(30-1)d$
$\displaystyle =100+29(5)$
$\displaystyle =100+145$
$\displaystyle =245$
$\displaystyle \therefore \text{The }30^{\text{th}}\text{ instalment is Rs. }245.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{A carpenter was hired to build }192\text{ window frames. On the first day, he}$
$\displaystyle \text{made }5\text{ frames, and each day thereafter he made }2\text{ more frames than on the previous}$
$\displaystyle \text{day. How many days did he take to finish the job?}$
$\displaystyle \text{Answer:}$
$\displaystyle a=5,\qquad d=2,\qquad S_n=192$
$\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]$
$\displaystyle 192=\frac{n}{2}\left[2(5)+(n-1)(2)\right]$
$\displaystyle 192=\frac{n}{2}(10+2n-2)$
$\displaystyle 192=n(n+4)$
$\displaystyle n^2+4n-192=0$
$\displaystyle (n-12)(n+16)=0$
$\displaystyle n=12\quad\text{or}\quad n=-16$
$\displaystyle \text{Since the number of days cannot be negative, }n=12.$
$\displaystyle \therefore \text{The carpenter took }12\text{ days to complete the job.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{The sum of the interior angles of a triangle is }180^\circ.\text{ Show that the}$
$\displaystyle \text{sums of the interior angles of polygons with }3,4,5,6,\ldots\text{ sides form an A.P.}$
$\displaystyle \text{Find the sum of the interior angles of a }21\text{-sided polygon.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The sum of the interior angles of a }3\text{-sided polygon is }180^\circ.$
$\displaystyle \text{The sum of the interior angles of a }4\text{-sided polygon is }360^\circ.$
$\displaystyle \text{The sum of the interior angles of a }5\text{-sided polygon is }540^\circ.$
$\displaystyle \therefore \text{The sequence is }180^\circ,360^\circ,540^\circ,\ldots$
$\displaystyle a=180^\circ,\qquad d=360^\circ-180^\circ=180^\circ$
$\displaystyle \therefore \text{These sums form an A.P.}$
$\displaystyle \text{Since the }3\text{-sided polygon corresponds to the first term, a }21\text{-sided polygon}$
$\displaystyle \text{corresponds to the }(21-2)^{\text{th}}=19^{\text{th}}\text{ term.}$
$\displaystyle a_{19}=a+(19-1)d$
$\displaystyle =180^\circ+18(180^\circ)$
$\displaystyle =180^\circ+3240^\circ$
$\displaystyle \therefore a_{19}=3420^\circ$
$\displaystyle \therefore \text{The sum of the interior angles of a }21\text{-sided polygon is }3420^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{In a potato race, }20\text{ potatoes are placed in a line at intervals of }4\text{ m,}$
$\displaystyle \text{the first potato being }24\text{ m from the starting point. A contestant brings the potatoes}$
$\displaystyle \text{back one at a time. Find the total distance covered.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The first round trip is }2\times24=48\text{ m.}$
$\displaystyle \text{Each successive round trip increases by }2\times4=8\text{ m.}$
$\displaystyle \therefore a=48,\qquad d=8,\qquad n=20$
$\displaystyle S_{20}=\frac{20}{2}\left[2(48)+(20-1)(8)\right]$
$\displaystyle =10(96+152)$
$\displaystyle =10\times248$
$\displaystyle =2480\text{ m}$
$\displaystyle \therefore \text{The contestant runs }2480\text{ m in bringing back all the potatoes.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{A man accepts a position with an initial salary of Rs. }5200\text{ per month.}$
$\displaystyle \text{His salary increases automatically by Rs. }320\text{ every month. Find:}$
$\displaystyle \text{i) his salary for the tenth month}\qquad\text{ii) his total earnings during the first year.}$
$\displaystyle \text{Answer:}$
$\displaystyle a=5200,\qquad d=320$
$\displaystyle \text{i) }a_{10}=a+(10-1)d$
$\displaystyle =5200+9(320)$
$\displaystyle =5200+2880$
$\displaystyle =8080$
$\displaystyle \therefore \text{The salary for the tenth month is Rs. }8080.$
$\displaystyle \text{ii) }S_{12}=\frac{12}{2}\left[2(5200)+(12-1)(320)\right]$
$\displaystyle =6(10400+3520)$
$\displaystyle =6(13920)$
$\displaystyle =83520$
$\displaystyle \therefore \text{The total earnings during the first year are Rs. }83520.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{A man saved Rs. }66000\text{ in }20\text{ years. In each year after the first, he}$
$\displaystyle \text{saved Rs. }200\text{ more than in the preceding year. How much did he save in the first year?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the amount saved in the first year be Rs. }a.$
$\displaystyle d=200,\qquad n=20,\qquad S_{20}=66000$
$\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]$
$\displaystyle 66000=\frac{20}{2}\left[2a+(20-1)(200)\right]$
$\displaystyle 66000=10(2a+3800)$
$\displaystyle 6600=2a+3800$
$\displaystyle 2a=2800$
$\displaystyle \therefore a=1400$
$\displaystyle \therefore \text{The man saved Rs. }1400\text{ in the first year.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{In a cricket tournament, }16\text{ teams participated. A sum of Rs. }8000\text{ is}$
$\displaystyle \text{to be distributed as prize money. The last-placed team receives Rs. }275,\text{ and the award}$
$\displaystyle \text{increases by an equal amount for each successive higher position. Find the prize money}$
$\displaystyle \text{received by the first-placed team.}$
$\displaystyle \text{Answer:}$
$\displaystyle a=275,\qquad n=16,\qquad S_{16}=8000$
$\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]$
$\displaystyle 8000=\frac{16}{2}\left[2(275)+(16-1)d\right]$
$\displaystyle 8000=8(550+15d)$
$\displaystyle 1000=550+15d$
$\displaystyle 15d=450$
$\displaystyle \therefore d=30$
$\displaystyle \text{The first-placed team receives the }16^{\text{th}}\text{ term.}$
$\displaystyle a_{16}=a+(16-1)d$
$\displaystyle =275+15(30)$
$\displaystyle =275+450$
$\displaystyle \therefore a_{16}=725$
$\displaystyle \therefore \text{The first-placed team receives Rs. }725.$
$\displaystyle \\$