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**Very Short Answer Type Questions****Q1. How many two-digit numbers are divisible by 7?****Sol.** Let 'n' two- digit numbers are divisible by 7.

So, the numbers are: 14, 21, 28, 35, 42, 49 ....................98.

This series form an A.P.

âˆ´ a = 14,d = 7,and an = 98

Now, using an = a + (n - 1) d, we get

98 = 14 + (n - 1) 7

â‡’ 98 = 14 + 7n - 7

â‡’ 98 = 7 + 7n or 7n

= 98 - 7 = 91 â‡’ n = 91/7 = 13

Thus, required two-digit numbers is 13**Q2. If the numbers x - 2, 4x - 1 and 5x + 2 are in A.P. Find the value of x.Sol. **âˆµ x - 2, 4x - 1 and 5x + 2 are in A.P.

âˆ´ (4x - 1) - (x - 2) = (5x + 2) - (4x - 1)

â‡’ 3x + 1 = x + 3

â‡’ 2x = 2 â‡’ x = 1

âˆ´ Using T

109 = 4 + (n - 1) 5

[âˆµ a = 4 and d = 9 - 4 = 5]

â‡’

â‡’ n = 21 + 1 = 22

Thus, the 22nd term is 109.**Q4. If a, (a - 2) and 3a are in A.P. then what is the value of a?****Sol.** âˆµ a, (a - 2) and 3a are in A.P.

âˆ´ (a - 2) - a = 3a - (a - 2)

â‡’ a - 2 - a = 3a - a + 2

â‡’ - 2 = 2a + 2

â‡’ 2a = - 2 - 2 = - 4

â‡’ a = -4/2 = -2

Thus, the required value of a is - 2.**Q5. How many terms are there in the A.P.?****7, 10, 13, ....., 151****Sol.** Here, a = 7, d = 10 - 7 = 3

Let there are n-terms.

âˆ´ T_{n} = a + (n - 1) d

â‡’ T_{51} = 7 + (n - 1) Ã— 3

â‡’

â‡’ 144/3 = n âˆ’ 1 â‡’ n = 48 + 1 = 49

i.e., n = 49**Q6. Which term of the A.P. 72, 63, 54, ..... is 0?****Sol.** Here, a = 72

d = 63 - 72 = - 9

Let nth term of this A.P. be 0

âˆ´ T_{n} = a + (n - 1) d

â‡’ 72 + (n - 1) Ã— (- 9) = 0

â‡’ (n - 1) = -72/-9 = 8

â‡’ n = 8 + 1 = 9

Thus the 9th term of the A.P. is 0.**Q7. The first term of an A.P. is 6 and its common difference is - 2. Find its 18th term.****Sol.** Using T_{n} = a + (n - 1) d, we have:

T_{18} = 6 + (18 - 1) Ã— (- 2)

= 6 + 17 Ã— (- 2)

= 6 - 34 = - 28

Thus, the 18th term is - 28.**Q8. The 4th term of an A.P. is 14 and its 12th term 70. What is its first term****Sol.** Let the first term = a

If â€˜dâ€™ is the common difference,

Then T_{4} = a + 3d = 14 ...(1)

And T_{12} = a + 11d = 70 ...(2)

Subtracting (1) from (2),

a + 11d - a - 3d = 70 - 14

â‡’ 8d = 56 â‡’ d = 56/8 = 7

âˆ´ From (1), a + 3 (7) = 14

â‡’ a + 21 = 14

â‡’ a = 14 - 21 = (- 7)

Thus, the first term is - 7.**Q9. Which term of A.P. 5, 2, - 1, - 4 ..... is - 40?****Sol.** Here, a = 5

d = 2 - 5 = - 3

Let nth term be - 40

âˆ´ T_{n} = a + (n - 1) d

â‡’ - 40 = 5 + (n - 1) Ã— (- 3)

â‡’ n = 15 + 1 = 16 i.e.,

i.e., The 16th term of the A.P. is - 40.**Q10. What is the sum of all the natural numbers from 1 to 100?****Sol.** We have:

1, 2, 3, 4, ....., 100 are in an A.P. such that

a = 1 and l = 100

âˆ´ S_{n} = n/2 [a + l]

â‡’ S_{100} =100/2 [1 + 100]

= 50 Ã— 101 = 5050.**Q11. For an A.P., the 8th term is 17 and the 14th term is 29. Find its common difference.****Sol. **Let the common difference = d and first term = a

âˆ´ T_{8} = a + 7d = 17 ...(1)

T_{14} = a + 13d = 29 ...(2)

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

a + 13d - a - 7d = 29 - 17

â‡’ 6d = 12

â‡’ d = 12/6 = 2

âˆ´ The required common difference = 2.**Q12. If the first and last terms of an A.P. are 10 and - 10. How many terms are there? Given that d = - 1.****Sol.** Let the required number of terms is n and

1st term a = 10

nth term T_{n} = - 10

Let common difference be d then using,

T_{n} = a + (n - 1) d, we have:

- 10 = 10 + (n - 1) Ã— (- 1)

â‡’ - 10 = 10 - n + 1

â‡’ - n + 1 = - 10 - 10 = - 20

â‡’ - n = - 20 - 1 = - 21

â‡’ n = 21**13. The nth term of an A.P. is (3n - 2) find its first term.Sol.** âˆµ T

âˆ´ T

â‡’ First term = 1

âˆ´ T

T

âˆ´ d = T

Thus the common difference is 2.

Sol.

âˆ´ T

T

âˆ´ a = 2

and d = T

= 9 - 2 = 7

Now T

[using T

= 2 + 99 Ã— 7

= 2 + 693 = 695.

d = 8 - 5 = 3

n = 12

Using S

we have: S

= 6 [10 + 33]

= 6 Ã— 43 = 258

âˆ´ T

T

â‡’ d = T

= 11 - 8 = 3

Thus, the common difference = 3.

Sol.

T

T

T

For an A.P., we have:

âˆ´ 2x - (x + 2) = 2x + 3 - 2x

â‡’ 2x - x - 2 = 2x + 3 - 2x

â‡’ x - 2 = 3

â‡’ x = 3 + 2 = 5

Thus, x = 5

âˆ´ T

And T

âˆµ d = T

âˆ´ d = 13 - 8 = 5

Thus, common difference = 5.

For an A.P., we have

T

i.e., 2x + k - x = 3x + 6 - (2x + k)

â‡’ x + k = 3x + 6 - 2x - k

â‡’ x + k = x + 6 - k

â‡’ k + k = x + 6 - x

â‡’ 2k = 6

â‡’ k = 6/2 = 3

For an A.P., T

âˆ´ (2k - 1) - (k + 9) = (2k + 7) - (2k - 1)

â‡’ 2k - 1 - k - 9 = 2k + 7 - 2k + 1

â‡’ k - 10 = 8

â‡’ k = 8 + 10

â‡’ k = 18

T

T

T

âˆµ For an A.P.,

T

âˆ´

â‡’

â‡’ 2a = 14/5

â‡’

Thus, a = 7/5

T

T

T

âˆµ For an A.P., we have:

T

â‡’ 7 - (2p - 1) = 3p - 7

â‡’ 7 - 2p + 1 = 3p - 7

â‡’ - 2p - 3p = - 7 - 1 - 7

â‡’ - 5p = - 15

â‡’ p = -15/-5 = 3

Thus, p = 3

T

T

T

For an A.P., we have:

T

â‡’ 13 - (2p + 1) = 5p - 3 - 13

â‡’ 13 - 2p - 1 = 5p - 16

â‡’ - 2p + 12 = 5p - 16

â‡’ - 2p - 5p = - 16 - 12 = - 28

â‡’ - 7p = - 28

â‡’

âˆ´ p = 4

âˆ´ T

T

âˆ´ d = T

= (- 1) - 3 = - 4

Thus, common difference = - 4

Sol.

âˆ´ T

T

â‡’ d = T

âˆ´ Common difference = 6.

Sol

Here,

âˆ´

Now, d = T

âˆ´ T

Thus, the next term of the A.P. is 5âˆš2 or .**Q28. The first term of an A.P. is p and its common difference is q. Find the 10th term.****Sol**. Here, a = p and d = q

âˆµ T_{n} = a + (n - 1) d

âˆ´ T_{10} = p + (10 - 1) q

= p + 9q

Thus, the 10th term is p + 9q.**Q29. (a) Find the next term of the A.P. ****(b) Find the tenth term of the sequence **** Sol.** (a) Here,

Now, d = T

Now, using T

T

Thus, the next term = âˆš32 .

(b) Here, T

d = 18 - 21 = - 3

Since Tn = a + (n - 1) d

â‡’ 0 = 21 + (n - 1) Ã— (- 3)

â‡’ - 3 (n - 1) = - 21

â‡’ (n - 1) = -21/-3 = 7

â‡’ n = 7 + 1 = 8

Thus, the 8th term of this A.P. will be 0.

d = 11 - 14 = - 3

Let the nth term be (- 1)

âˆ´ Using T

- 1 = 11 + (n - 1) Ã— (- 3)

â‡’ - 1 - 14 = - 3 (n - 1)

â‡’ - 15 = - 3 (n - 1)

âˆ´ n - 1 = -15/-3 = 5

â‡’ n = 5 + 1 = 6

Thus, -1 is the 6th term of the A.P.

âˆ´ a

â‡’ 49 = â€“11 + (n â€“ 1) Ã— 4

â‡’ n = 16

Since, n is an even number

âˆ´ There will be two middle terms, which are:

or 8th and 9th

Now, a

= â€“11 + 7 Ã— 4 = 17

a

= â€“11 + 8 Ã— 4 = 21

Thus, the values of the two middlemost terms are : 17 and 21.

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