Class 11: Arithmetic Progressions – Exercise 19.7 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: A man saved Rs. $\displaystyle 16500$ in ten years. In each year after the first, he saved Rs. $\displaystyle 100$ more than he did in the preceding year. How much did he save in the first year?

Answer:

Let the amount in Rs saved by the man in the first year $\displaystyle = x$

$\displaystyle \therefore x + ( x+100) + ( x+200) + \ldots + (x + 900) = 16500$

$\displaystyle \Rightarrow 10x + [ 100+ 200 + \ldots + 900] = 16500$

$\displaystyle \Rightarrow 10x + \frac{9}{2} \Big[ 2(100) + ( 9-1) (100) \Big] = 16500$

$\displaystyle \Rightarrow 10x + 4500 = 16500$

$\displaystyle \Rightarrow 10x = 12000$

$\displaystyle \Rightarrow x = 1200$

$\displaystyle \text{Therefore the man saved } 1200$ Rs. in the first year.

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Question 2: A man saves Rs. $\displaystyle 32$ during the first year, Rs. $\displaystyle 36$ in the second year and in this way he increases his savings by Rs. $\displaystyle 4$ every year. Find in what time his saving will be Rs. $\displaystyle 200$ .

Answer:

$\displaystyle \text{Given series is } 32, 36, 40, \ldots$

$\displaystyle \therefore a = 32, \ \ \ \ \ d = 36-32=4, \ \ \ \ \ \ S_n = 200$

$\displaystyle \text{We know, } S_n = \frac{n}{2} [ 2(32) + ( n-1)(4) ]$

$\displaystyle \Rightarrow 200 = n [ 32 + 2n-2]$

$\displaystyle \Rightarrow 2n^2 + 30 n - 200 = 0$

$\displaystyle \Rightarrow n^2 + 15n - 100 = 0$

$\displaystyle \Rightarrow (n-5)(n+20) = 0$

$\displaystyle \Rightarrow n = 5 \ or \ n = -20 \text{ (this is not possible) }$

$\displaystyle \text{Therefore man will save } 200$ Rs. in 5 year.

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Question 3: A man arranges to pay off a debt of Rs. $\displaystyle 3600$ by $\displaystyle 40$ annual instalments which form an arithmetic series. When $\displaystyle 30$ of the installments are paid, he dies leaving one-third of the debt unpaid, find the value of, the first installment.

Answer:

Let the first instalment be $\displaystyle a$ and the common difference be $\displaystyle d$ . Therefore

$\displaystyle a + ( a+d) + ( a+2d) + \ldots + (a+39d) = 3600$

$\displaystyle \Rightarrow 3600 = \frac{40}{2} [ 2(a) + ( 40-1) d]$

$\displaystyle \Rightarrow 3600 = 20 [ 2a +39 d]$

$\displaystyle \Rightarrow 2a + 39d = 180$ … … … … … i)

In the first $\displaystyle 30$ instalments, the man has paid $\displaystyle 2400$ Rs. Therefore

$\displaystyle 2400 = \frac{30}{2} [ 2(a) + ( 30-1) d ]$

$\displaystyle \Rightarrow 2400 = 15 [ 2a + 29 d]$

$\displaystyle \Rightarrow 2a + 29 d = 160$ … … … … … ii)

Solving i) and ii) we get

$\displaystyle 180 - 39d = 160 - 29 d \Rightarrow 20 = 10 d \Rightarrow d = 2$

Substituting in i) we get

$\displaystyle \therefore 2a = 180 - 39 ( 2)$

$\displaystyle \Rightarrow a = 51$

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Question 4: A manufacturer of radio sets produced $\displaystyle 600$ units in the third year and $\displaystyle 700$ units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find (i) the production in the first year (ii) the total product in $\displaystyle 7$ years and (iii) the product in the $\displaystyle 10^{th}$ year.

Answer:

$\displaystyle \text{Given } a_3 = 600 \ \ \ \ \ a_7 = 700$

$\displaystyle \text{We know, } a_n = a + ( n-1) d$

$\displaystyle \Rightarrow 600 = a+(3-1) d$

$\displaystyle \Rightarrow a + 2d = 600$ … … … … … i)

$\displaystyle \text{Also } 700 = a + (7-1) d$

$\displaystyle \Rightarrow a + 6d = 700$ … … … … … ii)

Solving i) and ii) we get

$\displaystyle 600-2d = 700 - 6d \Rightarrow 4d = 100 \Rightarrow d = 25$

$\displaystyle \therefore a = 700-6(25) = 700 - 150 = 550$

i) Production in first year $\displaystyle = 550$ units

ii) $\displaystyle S_7 = \frac{7}{2} [ 2 ( 550) + ( 7-1)(25) ] = \frac{7}{2} [ 1250] = 4375$

iii) $\displaystyle a_{10} = 550 + ( 10-1)(25) = 550 + 225 = 775$

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Question 5: There are $\displaystyle 25$ trees at equal distances of $\displaystyle 5$ meters in a line with a well, the distance of the well from the nearest tree being $\displaystyle 10$ meters. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.

Answer:

The series will be $\displaystyle 20, 30, 40, \ldots$

$\displaystyle \therefore a = 20 \ \ \ \ \ d = 30-20 = 10 \ \ \ \ \ n = 25$

$\displaystyle \text{We know, } S_n = \frac{n}{2} [ 2a+(n-1)d]$

$\displaystyle \therefore S_{25}= \frac{25}{2} [ 2(20) + ( 25-1)(10) ] = \frac{25}{2} [ 40 + 240] = 3560$

The total distance the gardener will cover in order to water all the trees $\displaystyle = 3560$ m.

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Question 6: A man is employed to count Rs. $\displaystyle 10710$ . He counts at the rate of Rs. $\displaystyle 180$ per minute for half an hour. After this, he counts at the rate of Rs. $\displaystyle 3$ less every minute than the preceding minute. Find the time taken by him to count the entire amount

Answer:

Amount of Rs. counted in first 30 minutes $\displaystyle = 180 \times 30 = 5400$ Rs.

Amount of money left for counting $\displaystyle = 10710 - 5400 = 5310$ Rs.

Therefore the series would be $\displaystyle (180-3), (180-6), (180-9) , \ldots$ or $\displaystyle 177, 174, 171, \ldots$

$\displaystyle \therefore a = 177 \ \ \ \ \ d= 174-177 = -3 \ \ \ \ \ S_n = 5310$

Let the time taken to count the rest of the money be n minutes

$\displaystyle \text{We know, } S_n = \frac{n}{2} [2a+(n-1)d]$

$\displaystyle \therefore 5310 = \frac{n}{2} [ 2 ( 177) + ( n-1) (-3) ]$

$\displaystyle \Rightarrow 5310 = \frac{n}{2} [ 354 - 3n + 3]$

$\displaystyle \Rightarrow 10620 = n [ 357 - 3n]$

$\displaystyle \therefore 3n^2 - 357n + 10620 = 0$

$\displaystyle (n-60)(n-59) = 0$

Therefore the time taken to count is $\displaystyle 59$ minutes.

Hence the total time taken to count the money $\displaystyle = 30 + 59 = 89$ minutes.

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Question 7: A piece of equipment cost a certain factory Rs. $\displaystyle 600,000$ . If it depreciates in value, $\displaystyle 15\%$ the first, $\displaystyle 13.5\%$ the next year, $\displaystyle 12\%$ the third year, and so on. What will be its value at the end of $\displaystyle 10$ years, all percentages applying to the original cost ?

Answer:

The percentage of depreciation of $\displaystyle 1^{st}, 2^{nd}, 3^{rd} \ldots$ years is $\displaystyle 15, 13.5, 12, \ldots$

$\displaystyle \therefore a = 15 \ \ \ \ d = 13.5 - 15 = - 1.5 \ \ \ \ \ n = 10$

Therefore total depreciation percentage is $\displaystyle S_{10}$

$\displaystyle S_{10} = \frac{10}{2} [ 2(15) + ( 10 - 1) ( -1.5) ] = 5[30-13.5] = 82.5$

$\displaystyle \text{Therefore the total depreciation } = \frac{82.5}{100} \times 600000 = 495000 \text{ Rs. }$

Therefore the value at the end of $\displaystyle 10^{th}$ year $\displaystyle = 600000-495000 = 105000$ Rs.

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Question 8: A farmer buys a used tractor for Rs. $\displaystyle 12000$ . He pays Rs. $\displaystyle 6000$ cash and agrees to pay the balance in annual instalments of Rs. $\displaystyle 500$ plus $\displaystyle 12\%$ interest on the unpaid amount. How much the tractor cost him?

Answer:

Cost of the tractor $\displaystyle = 12000 \text{ Rs. }$

Unpaid amount $\displaystyle = 12000 - 6000 = 6000 \text{ Rs. }$

The farmer pays $\displaystyle 500$ Rs. per month. This means that the farmer will pay the remaining principal in $\displaystyle 12$ instalments.

But he also has to pay the interest on the remaining principal at $\displaystyle 12\%$ .

Therefore the interest paid would be represented by a series i.e.

$\displaystyle 6000 \times \frac{12}{100}, 5500 \times \frac{12}{100}, 5000 \times \frac{12}{100}, \ldots \text{ or } 720, 660, 600, \ldots$

$\displaystyle \therefore a = 720 \ \ \ \ \ d = 660-720 =-60 \ \ \ \ \ n = 12$

$\displaystyle \therefore S_{12} = \frac{12}{2} [ 2(720) + ( 12-1) ( -60) ] = 6[ 1440 - 660] = 4680 \text{ Rs. }$

Therefore the total cost of the tractor $\displaystyle = 12000 + 4680 = 16680 \text{ Rs. }$

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Question 9: Shamshad Ati buys a scooter for Rs. $\displaystyle 22000$ . He pays Rs. $\displaystyle 4000$ cash and agrees to pay the balance in annual instalments of Rs. $\displaystyle 1000$ plus $\displaystyle 10\%$ interest on the unpaid amount. How much the scooter will cost him.

Answer:

Cost of the scooter $\displaystyle = 22000 \text{ Rs. }$

Unpaid amount $\displaystyle = 22000 - 4000 = 18000 \text{ Rs. }$

The farmer pays $\displaystyle 1000$ Rs. per month. This means that the farmer will pay the remaining principal in $\displaystyle 18$ instalments.

But he also has to pay the interest on the remaining principal at $\displaystyle 10\%$ .

Therefore the interest paid would be represented by a series i.e.

$\displaystyle 18000 \times \frac{10}{100}, 17000 \times \frac{10}{100}, 16000 \times \frac{10}{100}, \ldots \text{ or } 1800, 1700, 1600, \ldots$

$\displaystyle \therefore a = 1800 \ \ \ \ \ d = 1700-1800 =-100 \ \ \ \ \ n = 18$

$\displaystyle \therefore S_{18} = \frac{18}{2} [ 2(1800) + ( 18-1) ( -100) ] = 9[ 3600 - 1700] = 17100 \text{ Rs. }$

Therefore the total cost of the scooter $\displaystyle = 22000 + 17100 = 39100 \text{ Rs. }$

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Question 10: The income of a person is Rs. $\displaystyle 300,000$ in the first year and he receives an increase of Rs. $\displaystyle 10000$ to his income per year for the next $\displaystyle 19$ years. Find the total amount, he received in $\displaystyle 20$ years.

Answer:

$\displaystyle \text{Given } a = 300000 \ \ \ \ \ d = 10000 \ \ \ \ \ n = 20$

$\displaystyle \therefore S_{20} = \frac{20}{2} [ 2 ( 300000) + ( 20 -1) ( 10000) ] = 10 [ 600000 + 190000 ] = 7900000 \text{ Rs. }$

Therefore total amount received $\displaystyle = 7900000 \text{ Rs. }$

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Question 11: A man starts repaying a loan as the first installment of Rs. $\displaystyle 100$ . If he increases the instalments by Rs. $\displaystyle 5$ every month , what amount he will pay in the $\displaystyle 30^{th}$ instalment?

Answer:

$\displaystyle \text{Given } a = 100 \ \ \ \ \ d = 5 \ \ \ \ \ n = 30$

$\displaystyle \therefore a_{30} = 100 + ( 30 -1)(5) = 245$

Therefore the $\displaystyle 30^{th}$ instalment $\displaystyle = 245 \text{ Rs. }$

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Question 12: A carpenter was hired to build $\displaystyle 192$ window frames. The first day he made five frames and each day thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?

Answer:

$\displaystyle \text{Given } a = 5, \ \ \ \ d= 2 \ \ \ \ \ \ S_n = 192$

$\displaystyle S_n = \frac{n}{2} [ 2a + ( n-1) d]$

$\displaystyle \Rightarrow 192 = \frac{n}{2} [ 2(5) + ( n-1) (2) ]$

$\displaystyle \Rightarrow 192 = \frac{n}{2} [ 10 + 2n - 2]$

$\displaystyle \Rightarrow 192 = n ( 4 + n)$

$\displaystyle \Rightarrow n^2 + 4n - 192 = 0$

$\displaystyle \Rightarrow (n-12)(n+16) = 0$

$\displaystyle \Rightarrow n = 12 \ or \ n = -16 \text{ (this is not possible) }$

$\displaystyle \Rightarrow n = 12$

Therefore the carpenter will take $\displaystyle 12$ days to complete the task.

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Question 13: We know that the sum of the interior angles of a triangle is $\displaystyle 180^{\circ}$ . Show that the sums of the interior angles of polygons with $\displaystyle 3, 4, 5, 6, \ldots$ sides form an arithmetic progression. Find the sum of the interior angles for a $\displaystyle 21-$ sided polygon.

Answer:

We know that:

The sum of interior angles of a polygon with $\displaystyle 3$ sides $\displaystyle (a_1) = 180^{\circ}$

The sum of interior angles of a polygon with $\displaystyle 4$ sides $\displaystyle (a_2) = 360^{\circ}$

The sum of interior angles of a polygon with $\displaystyle 5$ sides $\displaystyle (a_3) = 540^{\circ}$

$\displaystyle \therefore a = 180^{\circ} \ \ \ \ d = 360^{\circ}-180^{\circ}= 180^{\circ}$

Since the sum of the interior angles with $\displaystyle 3$ sides $\displaystyle = a$ , then the sum of $\displaystyle 21$ sided polygon would be $\displaystyle a_{19}$

$\displaystyle \therefore a_{19} = 180^{\circ} + ( 19-1)(180^{\circ}) = 180^{\circ} + 3240^{\circ} = 3420^{\circ}$

The sum of the interior angles for a $\displaystyle 21-$ sided polygon $\displaystyle = 3420^{\circ}$

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Question 14: In a potato race $\displaystyle 20$ potatoes are placed in a line at intervals of $\displaystyle 4$ meters with, the first potato $\displaystyle 24$ meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Answer:

$\displaystyle \text{Given } a = 2 \times 24 = 48 \ \ \ \ d = 2 \times 4 = 8 \ \ \ \ n =20$

$\displaystyle \therefore S_{20} = \frac{20}{2} [ 2 ( 48) + ( 20-1) ( 4) ] = 10 [ 96 + 152] = 2480$

Therefore the contestant would run $\displaystyle 2480$ meters to bring back all the potatoes.

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Question 15: A man accepts a position with an initial salary of Rs. $\displaystyle 5200$ per month. It is understood that he will receive an automatic increase of Rs. $\displaystyle 320$ in the very next month and each month thereafter.

i) Find his salary for the tenth month

ii) What is his total earnings during the first year?

Answer:

$\displaystyle \text{Given } a = 5200 \ \ \ \ d = 320$

i) $\displaystyle a_{10} = 5200 + ( 10-1) ( 320 ) = 5200 + 2880 = 8080$

Therefore $\displaystyle 10^{th}$ month salary $\displaystyle = 8080 \text{ Rs. }$

ii) $\displaystyle S_{12} = \frac{12}{2} [ 2 ( 5200) + ( 12 -1) ( 320) ] = 6 [ 10400 + 3520 ] = 83520$

Therefore the total earnings in the first year $\displaystyle = 83520 \text{ Rs. }$

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Question 16: A man saved Rs. $\displaystyle 66000$ in $\displaystyle 20$ years. In each succeeding year after the first year, he saved Rs. $\displaystyle 200$ more than what he saved in the previous year. How much did he save in the first year?

Answer:

Let the savings for the first year $\displaystyle = a$

$\displaystyle \text{Given } d = 200, \ \ \ n = 20 \ \ \ S_{20} = 66000$

$\displaystyle \text{We know, } S_n = \frac{n}{2} [ 2(a) + ( n-1) (d) ]$

$\displaystyle 66000 = \frac{20}{2} [ 2(a)+(20-1)(200) ]$

$\displaystyle \Rightarrow 6600 = 2a + 3800$

$\displaystyle \Rightarrow 2a = 2800$

$\displaystyle \Rightarrow a = 1400$

$\displaystyle \text{Therefore the man saved } 1400 \text{ Rs. }$ in the first year.

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Question 17: In a cricket team tournament $\displaystyle 16$ teams participated. A sum of Rs. $\displaystyle 8000$ is to be awarded among themselves as prize money. If the last-placed team is awarded Rs. $\displaystyle 275$ in prize money and the award increases by the same amount for successive winning places, how much amount would the first team receive?