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**Ques 1: Fill in the blanks in the following table, given that â€˜aâ€™ is the first term, â€˜dâ€™ the common difference and a _{n} the nth term of the A.P.:**

a | d | n | an | |

(i) | 7 | 3 | 8 | â€¦ |

(ii) | -18 | â€¦ | 10 | 0 |

(iii) | â€¦ | -3 | 18 | -5 |

(iv) | -18.9 | 2.5 | â€¦ | 3.6 |

(v) | 3.5 | 0 | 105 | â€¦ |

**Sol: **(i) a_{n} = a + (n - 1) d

â‡’ a_{8} = 7 + (8 - 1) 3

= 7 + 7 Ã— 3

= 7 + 21

â‡’ a_{8} = 28

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

â‡’ a_{10} = - 18 + (10 - 1) d

â‡’ 0 = - 18 + 9d

â‡’ 9d = 18 â‡’ d = 18/9 = 2

âˆ´ d = 2

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

â‡’ - 5 = a + (18 - 1) Ã— (- 3)

â‡’ - 5 = a + 17 Ã— (- 3)

â‡’ - 5 = a - 51

â‡’ a = - 5 + 51 = 46

Thus, a = 46

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

â‡’ 3.6 = - 18.9 + (n - 1) Ã— 2.5

â‡’ (n - 1) Ã— 2.5 = 3.6 + 18.9

â‡’ (n - 1) Ã— 2.5 = 22.5

â‡’ n - 1 = 22.5/2.5 = 9

â‡’ n = 9 + 1 = 10

Thus, n = 10

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

â‡’ a_{n} = 3.5 + (105 - 1) Ã— 0

â‡’ a_{n} = 3.5 + 104 Ã— 0

â‡’ a_{n} = 3.5 + 0 = 3.5

Thus, a_{n} = 3.5**Ques ****2: Choose the correct choice in the following and justify:****(i) 30th term of the A.P.: 10, 7, 4, ...., is****(A) 97 ****(B) 77 ****(C) - 77 ****(D) - 87****(ii) 11th term of the A.P.: ****, ...., is(A) 28 **

âˆµ T

âˆ´ T

â‡’ T

â‡’ T

Thus, the correct choice is (C) - 77.

(ii) Here, a = - 3, n = 11 and

âˆ´T

â‡’

â‡’ T

Thus, the correct choice is (B) 22.

Let common difference = d

âˆ´ T

â‡’ T

â‡’ 26 = 2 + 2d

â‡’ 2d = 26 - 2 = 24

â‡’ d = 24/2 = 12

âˆ´ The missing term = a + d

= 2 + 12 =

(ii) Let the first term = a and common difference = d

Here, T

T

T

âˆ´T

â‡’ 2d = - 10

â‡’ d = -10/2 = -5

Now, a + d = 13 â‡’ a + (- 5) = 13

â‡’ a = 13 + 5 = 18

Thus, missing terms are a and a + 2d or 18 and 18 + (- 10) = 8

i.e., T

(iii) Here, a = 5 and T

since, T

â‡’ = 5 + 3d

â‡’ 3d =

â‡’ d =

âˆ´ The missing terms are:

T

T

(iv) Here, a = - 4 and T

âˆµ T

âˆ´T

â‡’ 6 = - 4 + 5d

â‡’ 5d = 6 + 4 = 10

â‡’ d = 10Â¸ 5 = 2

âˆ´T

T

T

T

âˆ´ The missing terms are

(v) Here, T

âˆ´T

T

â‡’ T

â‡’ 4d = - 60

â‡’ d = -60/4 = -15

âˆ´ a + d = 38 â‡’ a + (- 15) = 38

â‡’ a = 38 + 15 = 53

Now,

T

T

T

Thus, the missing terms are

Here, a = 3, â‡’ T

âˆ´ d = T

Now, T

â‡’ 78 = 3 + (n - 1) Ã— 5

â‡’ 78 - 3 = (n - 1) Ã— 5

â‡’ 75 = (n - 1) Ã— 5

â‡’ (n - 1) = 75Â¸ 5 = 15

â‡’ n = 15 + 1 = 16

Thus, 78 is the 16th term of the given A.P.

d = 13 - 7 = 6

Let the number of terms be n

âˆ´T

Now, T

â‡’ 7 + (n - 1) Ã— 6 = 205

â‡’ (n - 1) Ã— 6 = 205 - 7 = 198

â‡’ n - 1 = 198/6 = 33

âˆ´n = 33 + 1 = 34

Thus, the required number of terms is 34.

(ii) Here, a = 18

d =

Let the nth term = - 47

âˆ´T

â‡’

â‡’

â‡’

â‡’

â‡’ n - 1 = (- 13) Ã— (- 2) = 26

â‡’ n = 26 + 1 = 27

Thus, the required number of terms is 27.

a = 11

d = 8 - 11 = - 3

Let - 150 is the nth term of the given A.P.

âˆ´ T

â‡’ - 150 = 11 + (n - 1) Ã— (- 3)

â‡’ - 150 - 11 = (n - 1) Ã— (- 3)

â‡’ - 161 = (n - 1) Ã— (- 3)

â‡’

â‡’

But n should be a positive integer.

Thus, - 150 is not a term of the given A.P.

T

T

T

If the first term = a and the common difference = d.

Then,

a + (11 - 1) d = 38

â‡’ a + 10d = 38 ...(1)

and a + (16 - 1) d = 73

â‡’ a + 15d = 73 ...(2)

Subtracting (1) from (2), we get

(a + 15d) - (a + 10d) = 73 - 38

â‡’ 5d = 35

â‡’ d = 35/5 = 7

From (1),

a + 10 (7) = 38

â‡’ a + 70 = 38

â‡’ a = 38 - 70 = - 32

âˆ´T

â‡’ T

â‡’ T

â‡’ T

Thus, the 31st term is 178.

T

T

If first term = a and the common difference = d

âˆ´T

T

â‡’ T

â‡’ 47d = 94

â‡’ d = 94/47 = 2

From (1), we have

a + 2d = 12 â‡’ a + 2 (2) = 12

â‡’ a = 12 - 4 = 8

Now, T

= 8 + (28) Ã— 2

= 8 + 56 = 64

Thus, the 29th term is 64.

Sol:

T

âˆ´Using T

â‡’ T

T

Subtracting (1) from (2) we get

(a + 8d) - (a + 2d) = - 8 - 4

â‡’ 6d = - 12

â‡’ d = -12/6 = -2

Now, from (1), we have:

a + 2d = 4

â‡’ a + 2 (- 2) = 4

â‡’ a - 4 = 4

â‡’ a = 4 + 4 = 8

Let the nth term of the A.P. be 0.

âˆ´T

â‡’ 8 + (n - 1) Ã— (- 2) = 0

â‡’ (n - 1) Ã— - 2 = - 8

â‡’ n - 1 = -8/-2 = 4

â‡’ n = 4 + 1 = 5

Thus, the 5th term of the A.P. is 0.

Now, using

T

T

T

According to the condition,

T

â‡’ (a + 9d) + 7 = a + 16d

â‡’ a + 9d - a - 16d = - 7

â‡’- 7d = - 7 â‡’ d = 1

Thus, the common difference is 1.

d = 15 - 3 = 12

Using T

T

= 3 + 53 Ã— 12

= 3 + 636 = 639

Let a

âˆ´ a

â‡’ a

Now a

â‡’ 3 + (n - 1) Ã— 12 = 771

â‡’ (n - 1) Ã— 12 = 771 - 3 = 768

â‡’ (n - 1) = 768/12 = 64

â‡’ n = 64 + 1 = 65

Thus, 132 more than 54th term is the 65th term.

âˆ´T

And for the 2nd A.P., the first term = a'

âˆ´ T'

According to the condition, we have:

= T

â‡’ a + 99d - (a' + 99d) = 100

â‡’ a - a' = 100

Let, = T

âˆ´a + 999d - (a' + 999d) = x

â‡’ a - a' = x â‡’ x = 100

âˆ´The difference between the 1000th terms is 100.

The last such three digit number is 994.

âˆ´ The A.P. is 105, 112, 119, ....., 994

Here, a = 105 and d = 7

Let n be the required number of terms.

âˆ´ T

â‡’ 994 = 105 + (n - 1) Ã— 7

â‡’ (n - 1) Ã— 7 = 994 - 105 = 889

â‡’ (n - 1) = 889/7 = 147

â‡’ n = 127 + 1 = 128

Thus, 128 numbers of 3-digit are divisible by 7.

The multiple of 4 just below 250 is 248.

âˆ´ The A.P. is given by:

12, 16, 20, ....., 248

Here, a = 12 and d = 4

Let the number of terms = n

âˆ´ Using T

âˆ´ T

â‡’ 248 = 12 + (n - 1) Ã— 4

â‡’ (n - 1) Ã— 4 = 248 - 12 = 236

â‡’ n - 1 = 236/4 = 59

â‡’ n = 59 + 1 = 60

Thus, the required number of terms = 60.

âˆµ a = 63 and d = 65 - 63 = 2

âˆ´T

â‡’ T

âˆµ a = 3 and d = 10 - 3 = 7

âˆ´ T

â‡’ T

Now, according to the condition,

3 + (n - 1) Ã— 7 = 63 + (n - 1) Ã— 2

â‡’ (n - 1) Ã— 7 - (n - 1) Ã— 2 = 63 - 3

â‡’ 7n - 7 - 2n + 2 = 60

â‡’ 5n

â‡’ 5n = 60 + 5 = 65

â‡’ n = 65/5 = 13

Thus, the 13th terms of the two given A.Ps. are equal.

âˆ´ Using Tn = a + (n - 1) d, we have:

T

â‡’ a + 2d = 16 ...(1)

And T

According to the condition,

T

â‡’ (a + 6d) - (a + 4d) = 12

â‡’ a + 6d - a - 4d = 12

â‡’ 2d = 12

â‡’ d = 12/2 = 6 ...(2)

Now, from (1) and (2), we have:

a + 2 (6) = 16

â‡’ a + 12 = 16

â‡’ a = 16 - 12 = 4

âˆ´The required A.P. is

4, [4 + 6], [4 + 2 (6)], [4 + 3 (6)], .....

or 4, 10, 16, 22, .....

Here, d = 8 - 3 = 5

Since, the nth term before the last term is given by l - (n - 1) d,

âˆ´We have

20th term from the end = l - (20 - 1) Ã— 5

= 253 - 19 Ã— 5

= 253 - 95 = 158

And the common difference = d

âˆ´ Using Tn = a + (n - 1) d,

T

â‡’ (a + 3d) + (a + 7d) = 24

â‡’ 2a + 10d = 24

â‡’ a + 5d = 12 ...(1)

And T

â‡’ (a + 5d) + (a + 9d) = 44

â‡’ 2a + 14d = 44

â‡’ a + 7d = 22 ...(2)

Now, subtracting (1) from (2), we get

(a + 7d) - (a + 5d) = 22 - 12

â‡’ 2d = 10

â‡’ d = 10/2 = 5

âˆ´From (1), a + 5 Ã— 5 = 12

â‡’ a + 25 = 12

â‡’ a = 12 - 25 = - 13

Now, the first three terms of the A.P. are given by:

a, (a + d), (a + 2d)

or - 13, (- 13 + 5), [- 13 + 2 (5)]

or - 13, - 8, - 3

Say, in the nth year he gets â‚¹ 7000.

âˆ´ Using T

7000 = 5000 + (n - 1) Ã— 200

â‡’ (n - 1) Ã— 200 = 7000 - 5000 = 2000

â‡’ n - 1 = 2000/200 = 10

â‡’ n = 10 + 1 = 11

Thus, his income becomes â‚¹ 7000 in 11 years.

âˆµ In the nth week her savings become â‚¹ 20.75.

âˆ´T

âˆ´Using T

20.75 = 5 + (n - 1) Ã— (1.75)

â‡’ (n - 1) Ã— 1.75 = 20.75 - 5

â‡’ (n - 1) Ã— 1.75 = 15.75

â‡’ n - 1 =15.75/1.75 = 9

â‡’ n = 9 + 1 = 10

Thus, the required number of years = 10.

**Sum of First n Terms of an A.P.**

(i) If the first term of an A.P. is â€˜aâ€™ and the common difference is â€˜dâ€™ then the sum of its first n terms is given by:

(ii) If the last term of the A.P. is l then

Remember,

The sum of first n positive integers is given by:

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