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**Ques 6: The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum****Sol: **We have,

First term a = 17

Last term l = 350 = T_{n}

Common difference d = 9

Let the number of terms be â€˜nâ€™

âˆµ T_{n} = a + (n - 1) d

âˆ´ 350 = 17 + (n - 1) Ã— 9

â‡’ (n - 1) Ã— 9 = 350 - 17 = 333

â‡’ n - 1 = 333/9 = 37

â‡’ n = 37 + 1 = 38

Since, S_{n} = n/2 (a + l)

âˆ´ S_{38} = 38/2 (17 + 350)

= 19 (367) = 6973

Thus, n = 38 and S_{n} = 6973**Ques ****7: Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.****Sol: **Here, n = 22, T_{22} = 149 = l

d = 7

Let the first term of the A.P. be â€˜aâ€™.

âˆ´ T_{n} = a + (n - 1) d

â‡’ T_{n} = a + (22 - 1) Ã— 7

â‡’ a + 21 Ã— 7 = 149

â‡’ a + 147 = 149

â‡’ a = 149 - 147 = 2

Now, S_{22} = n/2 [a + l]

â‡’ S_{22} = 22/2 [2 + 149]

= 11 [151] = 1661

Thus S_{22} = 1661**Ques ****8: Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.****Sol:** Here, n = 51, T_{2} = 14 and T_{3} = 18

Let the first term of the A.P. be â€˜aâ€™ and the common difference is d.

âˆ´We have:

T_{2} = a + d â‡’ a + d = 14 ...(1)

T_{3} = a + 2d â‡’ a + 2d = 18 ...(2)

Subtracting (1) from 2, we get

a + 2d - a - d = 18 - 14

â‡’ d = 14

From (1), we get

a + d = 14 â‡’ a + 4 = 14

â‡’ a = 14 - 4 = 10

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

â‡’ S_{51} = 51/2 [(2 Ã— 10) + (51 - 1) Ã— 4]

= 51/2 [20 + 200]

= 51/2 [220]

= 51 Ã— 110 = 5610

Thus, the sum of 51 terms is 5610.**Ques ****9: If the sum of first 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of first n terms.****Sol:** Here, we have:

S_{7} = 49 and S_{17} = 289

Let the first term of the A.P. be â€˜aâ€™ and â€˜dâ€™ be the common difference, then

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

â‡’ S_{7} = 7/2 [2a + (7 - 1) d] = 49

â‡’ 7 (2a + 6d) = 2 Ã— 49 = 98

â‡’ 2a + 6d = 98/7 = 14

â‡’ 2 [a + 3d] = 14

â‡’ a + 3d = 14/2 = 7

â‡’ a + 3d = 7 ...(1)

Also, S_{17} = 17/2[2a + (17 - 1) d] = 289

â‡’ 17/2 (2a + 16d) = 289

â‡’ a + 8d = 289/17 = 17

â‡’ a + 8d = 17 ...(2)

Subtracting (1) from (2), we have:

a + 8d - a - 3d = 17 - 7

â‡’ 5d = 10

â‡’ d =10/5 = 2

Now, from (1), we have

a + 3 (2) = 7

â‡’ a = 7 - 6 = 1

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

= n/2 [2 Ã— 1 + (n - 1) Ã— 2]

= n/2 [2 + 2n - 2]

= n/2 [2_{n}]

= n Ã— n = n^{2}

Thus, the required sum of n terms = n^{2}.**Ques ****10: Show that a _{1}, a_{2}, ..., a_{n}, ... form an A.P. where a_{n} is defined as below:**

Putting n = 1, 2, 3, 4, ..... n, we get:

a

a

a

a

..... ..... .....

a

âˆ´ The A.P. in which a = 7 and d = 11 - 7 = 4 is:

7, 11, 15, 19, ....., (3 + 4n).

Now S

= 15/2 [14 + (14 Ã— 4)]

= 15/2 [14 + 56]

= 15/2[70]

= 15 Ã— 35 = 525

(ii) Here, a

Putting n = 1, 2, 3, 4, ....., n, we get

a

a

a

a

..... .....

âˆ´ The A.P. is:

4, - 1, - 6, - 11, ..... 9 - 5 (n) [having first term as 4 and d = - 1 - 4 = - 5]

âˆ´ S

= 15/2[8 + 14 Ã— (- 5)]

= 15/2[8 - 70]

**Ques 11: If the sum of the first n terms of an AP is 4n - n ^{2}, what is the first term (that is S_{1})? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.**

S

âˆ´ S

= 4 - 1 = 3 â‡’ First term = 3

S

= 8 - 4 = 4 â‡’ Sum of first two terms = 4

âˆ´ Second term (S

S

= 12 - 9 = 3 â‡’ Sum of first 3 terms = 3

âˆ´ Third term (S

S

= 36 - 81 = - 45

S

= 40 - 100 = - 60

âˆ´ Tenth term = S

Now, S

Also S

= 4n - 4 - [n

= 4n - 4 - n

= 6n - n

âˆ´ nth term = S

= [4n - n

= 4n - n

Thus,

S

S

S

a

6, 12, 18, ....., (6 Ã— 40).

And, these numbers are in A.P. such that

a = 6

d = 12 - 6 = 6 and a

âˆ´ S

= 20 [12 + 39 Ã— 6]

= 20 [12 + 234]

= 20 Ã— 246 = 4920

OR

S

S

= 20 Ã— 246 = 4920

Thus, the sum of first 40 multiples of 6 is 4920.

8, (8 Ã— 2), (8 Ã— 3), (8 Ã— 4), ....., (8 Ã— 15)

or 8, 16, 24, 32, ....., 120.

These numbers are in A.P., where

a = 8 and l = 120

âˆ´ S

= 15/2 [8 + 120]

Thus, the sum of first positive 15 multiples of 8 is 960.

1, 3, 5, 7, ....., 49

These numbers are in A.P. such that

a = 1 and l = 49

Here, d = 3 - 1 = 2

âˆ´ T

â‡’ 49 = 1 + (n - 1)

â‡’ 49 - 1 = (n - 1)

â‡’ (n - 1) = 48/2 = 24

âˆ´ n = 24 + 1 = 25

Now, S

Thus, the sum of odd numbers between 0 and 50 is 625.

1st day = â‚¹ 200

2nd day = â‚¹ 250

3rd day = â‚¹ 300

...............

...............

Now, 200, 250, 300, ..... are in A.P. such that

a = 200, d = 250 - 200 = 50

âˆ´ S

S

= 15 [400 + 29 Ã— 50]

= 15 [400 + 1450]

= 15 Ã— 1850 = 27,750

Thus, penalty for the delay for 30 days is â‚¹ 27,750.

Let the first prize = a

âˆ´ 2nd prize = (a - 20)

3rd prize = (a - 40)

4th prize = (a - 60)

........................................

Thus, we have, first term = a

Common difference = - 20

Number of prizes, n = 7

Sum of 7 terms Sn = 700

Since, S

â‡’ 700 = 7/2 [2 (a) + (7 - 1) Ã— (- 20)]

â‡’ 700 = 7/2 [2a + (6 Ã— - 20)]

â‡’

â‡’ 200 = 2a - 120

â‡’ 2a = 200 + 120 = 320

â‡’ a = 320/2 = 160

Thus, the values of the seven prizes are:

â‚¹ 160, â‚¹ (160 - 20), â‚¹ (160 - 40), â‚¹ (160 - 60), â‚¹ (160 - 80), â‚¹ (160 - 100) and

â‚¹ (160 - 120)

â‡’ â‚¹ 160, â‚¹ 140, â‚¹ 120, â‚¹ 100, â‚¹ 80, â‚¹ 60 and â‚¹ 40.

âˆµ Each class has 3 sections.

âˆ´ Number of plants planted by class I = 1 Ã— 3 = 3

Number of plants planted by class II = 2 Ã— 3 = 6

Number of plants planted by class III = 3 Ã— 3 = 9

Number of plants planted by class IV = 4 Ã— 3 = 12

.......................................................................................................

Number of plants planted by class XII = 12 Ã— 3 = 36

The numbers 3, 6, 9, 12, ..........., 36 are in A.P.

Here, a = 3 and d = 6 - 3 = 3

âˆµ Number of classes = 12

i.e., n = 12

âˆ´ Sum of the n terms of the above A.P., is given by

S_{12} = 12/2 [2 (3) + (12 - 1) 3]

= 6 [6 + 11 Ã— 3]

= 6 [6 + 33]

= 6 Ã— 39 = 234

Thus, the total number of trees = 234.**Ques ****18: A spiral is made up of successive semi-circles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ..... as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semi-circles? (Take Ï€ =22/7)****[Hint: Length of successive semi-circles is l _{1}, l_{2}, l_{3}, l_{4}, ... with centres at A, B, A, B, ..., respectively.]**

= 1/2 (2Ï€r)

= Ï€r

âˆ´ l

l

l

l

...... ............... ......................

l

Now, length of the spiral

= l

= 0.5 Ï€ [1 + 2 + 3 + 4 + ..... + 13] cm ...(1)

âˆµ 1, 2, 3, 4, ....., 13 are in A.P. such that

a = 1 and l = 13

âˆ´ S

âˆ´ From (1), we have:

Total length of the spiral

= 1.5 Ï€ [91] cm

**Sol:** We have:

The number of logs:

1st row = 20

2nd row = 19

3rd row = 18

obviously, the numbers

20, 19, 18, ....., are in A.P. such that

a = 20

d = 19 - 20 = - 1

Let the numbers of rows be n.

âˆ´ S_{n} = 200

Now, using, Sn = [2a + (n - 1) d], we get

S_{n} = n/2 [2 (20) + (n - 1) Ã— (- 1)]

â‡’ 200 = n/2 [40 - (n - 1)]

â‡’ 2 Ã— 200 = n Ã— 40 - n (n - 1)

â‡’ 400 = 40n - n^{2} + n

â‡’ n^{2} - 41n + 400 = 0

â‡’ n^{2} - 16n - 25n + 400 = 0

â‡’ n (n - 16) - 25 (n - 16) = 0

â‡’ (n - 16) (n - 25) = 0

Either

â‡’ n - 16 = 0 â‡’ n = 16

or n - 25 = 0 â‡’ n = 25

T_{n} = 0 â‡’ a + (n - 1) d = 0 â‡’ 20 + (n - 1) Ã— (- 1) = 0

â‡’ n - 1 = 20 â‡’ n = 21

i.e., 21st term becomes 0

âˆ´ n = 25 is not required.

Thus, n = 16

âˆ´ Number of rows = 16

Now, T_{16} = a + (16 - 1) d

= 20 + 15 Ã— (- 1)

= 20 - 15 = 5

âˆ´ Number of logs in the 16th (top) row is 5.**Ques ****20: In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.**

**A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?****[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 Ã— 5 + 2 Ã— (5 + 3)]****Sol:** Here, number of potatoes = 10

The up-down distance of the bucket:

From the 1st potato = [5 m] Ã— 2 = 10 m

From the 2nd potato = [(5 + 3) m] Ã— 2 = 16 m

From the 3rd potato = [(5 + 3 + 3) m] Ã— 2 = 22 m

From the 4th potato = [(5 + 3 + 3 + 3) m] Ã— 2 = 28 m

.................................. ...........................

âˆµ 10, 16, 22, 28, ..... are in A.P. such that

a = 10 and d = 16 - 10 = 6

âˆ´ Using S_{n }= n/2 [2a + (n - 1) d], we have:

S_{10} = 10/2[2 (10) + (10 - 1) Ã— 6]

= 5 [20 + 9 Ã— 6]

= 5 [20 + 54]

= 5 [74]

= 5 Ã— 74 = 370

Thus, the sum of above distances = 370 m.

â‡’ The competitor has to run a total distance of 370 m.

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