RD Chapter 19- Arithmetic Progressions Ex-19.1 |
RD Chapter 19- Arithmetic Progressions Ex-19.2 |
RD Chapter 19- Arithmetic Progressions Ex-19.3 |
RD Chapter 19- Arithmetic Progressions Ex-19.4 |
RD Chapter 19- Arithmetic Progressions Ex-19.5 |
RD Chapter 19- Arithmetic Progressions Ex-19.6 |

**Answer
1** :

Given: A mansaved ₹16500 in ten years

Let ₹ x be his savingsin the first year

His savings increasedby ₹ 100 every year.

So,

A.P will be x, 100 +x, 200 + x………………..

Where, x is first termand

Common difference, d =100 + x – x = 100

We know, S_{n} isthe sum of n terms of an A.P

By using the formula,

S_{n} =n/2 [2a + (n – 1)d]

where, a is firstterm, d is common difference and n is number of terms in an A.P.

Given:

S_{n} =16500 and n = 10

S_{10} =10/2 [2x + (10 – 1)100]

16500 = 5{2x + 9(100)}

16500 = 5(2x + 900)

16500 = 10x + 4500

-10x = 4500 – 16500

–10x = –12000

x = 12000/10

= 1200

Hence, his saving inthe first year is ₹ 1200.

**Answer
2** :

Given:

First year savings is₹ 32

Second year savings is₹ 36

In this process heincreases his savings by ₹ 4 every year

Then,

A.P. will be 32, 36,40,………

Where, 32 is firstterm and common difference, d = 36 – 32 = 4

We know, S_{n} isthe sum of n terms of an A.P

By using the formula,

S_{n} =n/2 [2a + (n – 1)d]

where, a is firstterm, d is common difference and n is number of terms in an A.P.

Given:

S_{n} =200, a = 32, d = 4

S_{n} =n/2 [2a + (n – 1)d]

200 = n/2 [2(32) +(n-1)4]

200 = n/2 [64 + 4n –4]

400 = n [60 + 4n]

400 = 4n [15 + n]

400/4 = n [15 + n]

100 = 15n + n^{2}

n^{2} +15n – 100 = 0

n^{2} +20n – 5n – 100 = 0

n (n + 20) – 5 (n +20) = 0

(n + 20) – 5 (n + 20)= 0

(n + 20) (n – 5) = 0

n = -20 or 5

n = 5 [Since, n is apositive integer]

Hence, the manrequires 5 days to save ₹ 200

**Answer
3** :

Given:

40 annual instalmentswhich form an arithmetic series.

Let the firstinstalment be ‘a’

S_{40} =3600, n = 40

By using the formula,

S_{n} =n/2 [2a + (n – 1)d]

3600 = 40/2 [2a +(40-1)d]

3600 = 20 [2a + 39d]

3600/20 = 2a + 39d

2a + 39d – 180 = 0 …..(i)

Given:

Sum of first 30 termsis paid and one third of debt is unpaid.

So, paid amount = 2/3× 3600 = ₹ 2400

S_{n} =2400, n = 30

By using the formula,

S_{n} =n/2 [2a + (n – 1)d]

2400 = 30/2 [2a +(30-1)d]

2400 = 15 [2a + 29d]

2400/15 = 2a + 29d

2a + 29d -160 = 0 ….(ii)

Now, let us solveequation (i) and (ii) by substitution method, we get

2a + 39d = 180

2a = 180 – 39d … (iii)

Substitute the valueof 2a in equation (ii)

2a + 29d – 160 = 0

180 – 39d + 29d – 160= 0

20 – 10d = 0

10d = 20

d = 20/10

= 2

Substitute the valueof d in equation (iii)

2a = 180 – 39d

2a = 180 – 39(2)

2a = 180 – 78

2a = 102

a = 102/2

= 51

Hence, value of firstinstallment ‘a’ is ₹ 51

A manufacturer of the radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find

(і) the production in the first year

(іі) the total product in the 7 years and

(ііі) the product in the 10th year.

**Answer
4** :

Given:

600 and 700 radio setsunits are produced in third and seventh year respectively.

a_{3} =600 and a_{7} = 700

(і) The production in thefirst year

We need to find theproduction in the first year.

Let first yearproduction be ‘a’

So the AP formed is,a, a+x, a+2x, ….

By using the formula,

a_{n} = a+ (n-1)d

a_{3} = a+ (3-1)d

600 = a + 2d …. (i)

a_{7} = a+ (7-1)d

700 = a + 6d

a = 700 – 6d…. (ii)

Substitute value of ain (i) we get,

600 = a + 2d

600 = 700 – 6d + 2d

700 – 600 = 4d

100 = 4d

d = 100/4

= 25

Now substitute valueof d in (ii) we get,

a = 700 – 6d

= 700 – 6(25)

= 700 – 150

= 550

∴ The production in thefirst year, ‘a’ is 550

(іі) the total product inthe 7 years

We need to find thetotal product in 7 years i.e. is S_{7}

By using the formula,

S_{n} =n/2 [2a + (n-1)d]

n = 7, a = 550, d = 25

S_{7} =7/2 [2(550) + (7-1)25]

= 7/2 [1100 + 150]

= 7/2 [1250]

= 7 [625]

= 4375

∴ The total product inthe 7 years is 4375.

(ііі) the product in the 10^{th} year.

We need to find theproduct in the 10^{th} year i.e. a_{10}

By using the formula,

a_{n} = a+ (n-1)d

n = 10, a = 550, d =25

a_{10} =550 + (10-1)25

= 550 + (9)25

= 550 + 225

= 775

∴ The product in the 10^{th} yearis 775.

**Answer
5** :

Given: totaltrees are 25 and equal distance between two adjacent trees are 5 meters

We need to find thetotal distance the gardener will cover.

As gardener is comingback to well after watering every tree:

Distance covered bygardener to water 1^{st} tree and return back to the initialposition is 10m + 10m = 20m

Now, distance betweenadjacent trees is 5m.

Distance covered byhim to water 2^{nd} tree and return back to the initial positionis 15m + 15m = 30m

Distance covered bythe gardener to water 3^{rd} tree return back to the initialposition is 20m + 20m = 40m

Hence distance coveredby the gardener to water the trees are in A.P

A.P. is 20, 30, 40………upto 25 terms

Here, first term, a =20, common difference, d = 30 – 20 = 10, n = 25

We need to find S_{25} whichwill be the total distance covered by gardener to water 25 trees.

So by using theformula,

S_{n} =n/2 [2a + (n – 1)d]

S_{25} =25/2 [2(20) + (25-1)10]

= 25/2 [40 + (24)10]

= 25/2 [40 + 240]

= 25/2 [280]

= 25 [140]

= 3500

∴ The total distancecovered by gardener to water trees all 25 trees is 3500m.

**Answer
6** :

Given: Amount tobe counted is ₹ 10710

We need tofind time taken by man to count the entire amount.

He counts at the rateof ₹ 180 per minute for half an hour or 30 minutes.

So, Amount to becounted in an hour = 180 × 30 = ₹ 5400

Amount left = 10710 –5400 = ₹ 5310

S_{n} =5310

After an hour, rate ofcounting is decreasing at ₹ 3 per minute. This rate will form an A.P.

A.P. is 177, 174,171,……

Here a = 177 and d =174 – 177 = –3

By using the formula,

S_{n} =n/2 [2a + (n – 1)d]

5310 = n/2 [2(177) +(n-1) (-3)]

5310 = n/2 [354 -3n +3]

5310 × 2 = n [357 –3n]

10620 = 357n – 3n^{2}

10620 = 3n(119 – n)

10620/3 = n(119 – n)

3540 = 119n – n^{2}

n^{2} –119n + 3540 = 0

n^{2} –59n – 60n + 3540 = 0

n(n – 59) – 60(n – 59)= 0

(n – 59) (n – 60) = 0

n = 59 or 60

We shall considervalue of n = 59. Since, at 60^{th} min he will count ₹ 0

∴ The total time takenby him to count the entire amount = 30 + 59 = 89 minutes.

**Answer
7** :

Given: A piece ofequipment cost a certain factory is ₹ 600,000

We need to find thevalue of the equipment at the end of 10 years.

The price of equipmentdepreciates 15%, 13.5%, 12% in 1^{st}, 2^{nd}, 3^{rd} yearand so on.

So the A.P. will be15, 13.5, 12,…………… up to 10 terms

Here, a = 15, d = 13.5– 15 = –1.5, n = 10

By using the formula,

S_{n} =n/2 [2a + (n – 1)d]

S_{10} =10/2 [2(15) + (10-1) (-1.5)]

= 5 [30 + 9(-1.5)]

= 5 [30 – 13.5]

= 5 [16.5]

= 82.5

The value of equipmentat the end of 10 years is = [100 – Depreciation %]/100 × cost

= [100 – 82.5]/100 ×600000

= 175/10 × 6000

= 175 × 600

= 105000

∴ The value ofequipment at the end of 10 years is ₹ 105000.

**Answer
8** :

Given: Price ofthe tractor is ₹12000.

We need tofind the total cost of the tractor if he buys it in installments.

Total price = ₹ 12000

Paid amount = ₹ 6000

Unpaid amount = ₹12000 – 6000 = ₹ 6000

He pays remaining ₹6000 in ‘n’ number of installments of ₹ 500 each.

So, n = 6000/500 = 12

Cost incurred by himto pay remaining 6000 is

The AP will be:

(500 + 12% of 6000) +(500 + 12% of 5500) + … up to 12 terms

500 × 12 + 12% of(6000 + 5500 + … up to 12 terms)

By using the formula,

S_{n} =n/2 [2a + (n – 1)d]

n = 12, a = 6000, d =-500

S_{12} =500×12 + 12/100 × 12/2 [2(6000) + (12-1) (-500)]

= 6000 + 72/100 [12000+ 11 (-500)]

= 6000 + 72/100 [12000– 5500]

= 6000 + 72/100 [6500]

= 6000 + 4680

= 10680

Total cost = 6000 +10680

= 16680

∴ The total cost of thetractor if he buys it in installment is ₹ 16680.

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