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**Ques 1: Find the sum of the following APs:****(i) 2, 7, 12, ..., to 10 terms. ****(ii) - 37, - 33, - 29, ..., to 12 terms.****(iii) 0.6, 1.7, 2.8, ..., to 100 terms. ****(iv)**** , ..., to 11 terms.****Sol:** (i) Here, a = 2

d = 7 - 2 = 5

n = 10

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

âˆ´ S_{10} = [2 Ã— 2 + (10 - 1) Ã— 5]

â‡’ S_{10} = 5 [4 + 9 Ã— 5]

â‡’ S_{10 }= 5 [49] = 245

Thus, the sum of first 10 terms is 245.

(ii) We have:

a = - 37

d = - 33 - (- 37) = 4

n = 12

âˆ´ S_{n} = n/2 [2a + (n - 1) d]

â‡’ S_{12} = 12/2 [2 (- 37) + (12 - 1) Ã— 4]

= 6 [- 74 + 11 Ã— 4]

= 6 [- 74 + 44]

= 6 Ã— [- 30] = - 180

Thus, sum of first 12 terms = -180.

(iii) Here, a = 0.6

d = 1.7 - 0.6 = 1.1

n = 100

âˆ´ S_{n} = n/2 [2a + (n - 1) d]

S_{100} = 100/2 [2 (0.6) + (100 - 1) Ã— 1.1]

= 50 [1.2 + 99 Ã— 1.1]

= 50 [1.2 + 108.9]

= 50 [110.1]

= 5505

Thus, the required sum of first 100 terms is 5505.

(iv) Here,

a = 1/15

d =

n = 11

âˆ´ S_{n} = n/2 [2a + (n - 1) d]

S_{11} =

Thus, the required sum of first 11 terms = 33/20.**Ques 2: Find the sums given below:****(i) **** ****(ii) 34 + 32 + 30 + ... + 10****(iii) - 5 + (- 8) + (- 11) + ... + (- 230)****Sol:** (i) Here, a = 7

l = 84

Let n be the number of terms

âˆ´ T_{n} = a + (n - 1) d

â‡’

â‡’

â‡’ n = 22 + 1 = 23

Now,

â‡’

Thus, the required sum =

(ii) Here,a = 34

d = 32 - 34 = - 2

l = 10

Let the number of terms be n

âˆ´ T_{n} = 10

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

â‡’ 10 = 34 + (n - 1) Ã— (- 2)

â‡’ (n - 1) Ã— (- 2) = 10 - 34 = - 24

â‡’

â‡’ n = 13

â‡’

Now,

â‡’

=

=

=

= 13 Ã— 22 = 286

OR

S_{13} = n/2 (a + l)

Thus, the required sum is 286.

(iii) Here, a = - 5

d = - 8 - (- 5) = - 3

l = - 230

Let n be the number of terms.

âˆ´ T_{n} = - 230

â‡’ - 230 = - 5 + (n - 1) Ã— (- 3)

â‡’ (n - 1) Ã— (- 3) = - 230 + 5 = - 225

â‡’ n - 1 = -225/-3 = 75

â‡’ n = 75 + 1 = 76

Now,

= 38 Ã— (- 235)

= - 8930

âˆ´ The required sum = - 8930.**Ques ****3: In an AP:****(i) given a = 5, d = 3, a _{n} = 50, find n and S_{n}.**

âˆµ a

âˆ´ 50 = 5 + (n - 1) Ã— 3

â‡’ 50 - 5 = (n - 1) Ã— 3

â‡’ (n - 1) Ã— 3 = 45

â‡’ (n - 1) = 45/3 =15

â‡’ n = 15 + 1 = 16

Now S

= 16/2 (5 + 50)

= 8 (55) = 440

Thus, n = 16 and S

(ii) Here, a = 7 and a

âˆ´ a

â‡’ 35 = 7 + (13 - 1) d

â‡’ 35 - 7 = 12d

â‡’ 28 = 12d

â‡’ d = 28/12 = 7/3

Now, using

S

S

S

(iii) Here, a

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

Now a

â‡’ 37 = a + 11d

â‡’ 37 = a + 11 Ã— 3

â‡’ 37 = a + 33

â‡’ a = 37 - 33 = 4

Now, S

â‡’ S

â‡’ S

Thus, a = 4 and S

(iv) Here, a

S

Let first term of the AP be â€˜aâ€™ and the common difference = d

âˆ´ a

â‡’ a + 2d = 15 ...(1)

Again S

â‡’ S

â‡’ 125 = 5 [2a + 9d]

â‡’ 2a + 9d = 125/5 = 25

â‡’ 2a + 9d = 25 ...(2)

Multiplying (1) by 2 and subtracting (2) from it, we get

2 [a + 2d = 15] - [2a + 9d = 25]

â‡’ 2a + 4d - 2a - 9d = 30 - 25

â‡’ - 5d = 5

â‡’ d =5/-5 = -1

âˆ´ From (1), a + 2 (- 1) = 15 â‡’ a = 15 + 2 â‡’ a = 17

Now, a

= 17 + 9 Ã— (- 1)

= 17 - 9 = 8

Thus, d = - 1 and a

(v) Here, d = 5, S

Let the first term of the AP is â€˜aâ€™.

âˆ´ S

â‡’ 75 = 9/2 [2a + 40]

â‡’

â‡’ 50/3 = 2a + 40

â‡’

â‡’

Now, a

=

=

Thus,

(vi) Here, a = 2, d = 8 and S

âˆµ S

âˆ´ 90 = n/2 [2 Ã— 2 + (n - 1) Ã— 8]

â‡’ 90 Ã— 2 = 4n + n (n - 1) Ã— 8

â‡’ 180 = 4n + 8n

â‡’ 180 = 8n

â‡’ 45 = 2n

â‡’ 2n

â‡’ 2n

â‡’ 2n (n - 5) + 9 (n - 5) = 0

â‡’ (2n + 9) (n - 5) = 0

âˆ´ Either

or n - 5 = 0 â‡’ n = 5

But is not required. âˆ´n = 5

Now, a

â‡’ a

= 2 + 32 = 34

Thus, n = 5 and a

(vii) Here, a = 8, a

Let the common difference = d

Now, S

â‡’ 210 = n/2 (a + l)

â‡’ 210 = n/2 (8 + 62) =

âˆ´ n = 210/35 = 6

Again a

â‡’ 62 = 8 + (6 - 1) Ã— d

â‡’ 62 - 8 = 5d

â‡’ 54 = 5d â‡’ d = 54/5

Thus, n = 6 and d = 54/5 .

(viii) Here, a

Let the first term be â€˜aâ€™.

âˆµ a

âˆ´ a + (n - 1) 2 = 4

â‡’ a + 2n - 2 = 4

â‡’ a = 4 - 2n + 2

â‡’ a = 6 - 2n ...(1)

Also S

â‡’ n/2 (a + l) = - 14

â‡’ n/2 (a + 4) = - 14

â‡’ n (a + 4) = - 28 ...(2)

Substituting the value of a from (1) into (2),

n [6 - 2n + 4] = - 28

â‡’ n [10 - 2n] = - 28

â‡’ 2n [5 - n] = - 28

â‡’ n (5 - n) = - 14 [Dividing throughout by 2]

â‡’ 5n - n

â‡’ n

â‡’ n

â‡’ n (n - 7) + 2 (n - 7) = 0

â‡’ (n - 7) (n + 2) = 0

âˆ´ Either n - 7 = 0 â‡’ n = 7

or n + 2 = 0 â‡’ n = - 2

But n cannot be negative,

âˆ´ n = 7

Now, from (1), we have

a = 6 - 2 Ã— 7 â‡’ a = - 8

Thus, a = - 8 and n = 7

(ix) Here, a = 3, n = 8 and S

Let the common difference = d.

âˆµ S

âˆ´ 192 = 8/2 [2 (3) + (8 - 1) d]

â‡’ 192 = 4 [6 + 7d]

â‡’ 192 = 24 + 28d

â‡’ 28d = 192 - 24 = 168

â‡’ d = 168/28 =6

Thus, d = 6.

(x) Here, l = 28 and S

Let the first term be â€˜aâ€™.

Then S

â‡’ S

â‡’ 144 = 9/2 (a + 28)

â‡’

â‡’ a = 32 - 28 = 4

Thus, a = 4.

d = 17 - 9 = 8

S

âˆµ S

âˆ´ n/2 [(2 Ã— 9) + (n - 1) Ã— 8] = 636

â‡’ n [18 + (n - 1) Ã— 8] = 1272

â‡’ n (8n + 10) = 1272

â‡’ 8n

â‡’ 4n

â‡’ 4n

â‡’ n (4n + 53) - 12 (4n + 53) = 0

â‡’ (n - 12) (4n + 53) = 0

â‡’ n = 12 and

Rejecting , we have n = 12.

l = 45 = T

S

âˆµ T

âˆ´ 45 = 5 + (n - 1) d

â‡’ (n - 1) d = 45 â€“ 5

â‡’ (n - 1) d = 40 ...(1)

Also S

â‡’ 400 = n/2 (5 + 45)

â‡’ 400 Ã— 2 = n Ã— 50

â‡’

From (1), we get

(16 - 1) d = 40

â‡’ 15d = 40

â‡’ d = 40/15 = 8/3

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