The document Short Answer Type Questions (Part-1) - Arithmetic Progressions Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.

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** Note:** For an A.P. with the 1st term and common difference ‘a’ and ‘d’ respectively, we have :

(a) n

⇒ n

(b) If ‘l’ is the last term of the A.P., then

n

⇒ nth term from the end = l + (n – 1) (–d)

Now, Using T

T

T

[∵ a + 8d = 0]

T

= (a + 8d) + 20d

= 0 + 20d = 20d [∵ a + 8d = 0]

= 2 × (10d) = 2 (T19) [∵ T

⇒ T

Thus, the 29th term of the A.P. is double of its 19th term.

∵ T

∴ T

T

But it is given that

T

∴ T

⇒ First term, a = 7

Also, T

∴d = T

Now, using S

we get

S

Thus, the sum of the first 15 terms = 525.

3, 15, 27, 39, .....

∴ a = 3

d = 15 - 3 = 12

∴ Using, T

T

= 3 + (52 × 12)

= 3 + 624 = 627

Now, T

Let the required term be T

∴ T

or a + (n - 1) d = 747

∴ 3 + (n - 1) × 12 = 747

⇒ (n - 1) × 12 = 747 - 3 = 744

⇒ n - 1 = 744/12 = 62

⇒ n = 62 + 1 = 63

Thus, the 63rd term of the given A.P. is 120 more than its 53rd term.

Using T

T

⇒ 31 = a + 9d ...(1)

Also T

⇒ 66 = a + 14d ...(2)

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

a + 14d - a - 9d = 66 - 31

⇒ 5d = 35

⇒ d = 35/5 = 7

∴ From (1), a + 9d = 31

⇒ a + 9 (7) = 31

⇒ a + 63 = 31

⇒ a = 31 - 63

⇒ a = - 32

Now, T

= - 32 + 30 (7)

= - 32 + 210 = 178

Thus, the 31st term of the given A.P. is 178.

And the common difference = d

∴ Using T

∴ T

⇒ 37 = a + 7d ...(1)

Also T

And T

According to the question,

T

⇒ a + 14d = a + 11d + 15

⇒ a - a + 14d - 11d = 15

⇒ 3d = 15 ⇒ d = 15/3 = 5

From (1), we have:

a + 7 (5) = 37

⇒ a + 35 = 37

⇒ a = 37 - 35 = 2

Since an A.P. is given by :

a, a + d, a + 2d, a + 3d, ....

∴ The required A.P. is given by 2, 2 + 5, 2 + 2(5),... 2, 7, 12, ...

Now, using S

∴ S

= 15/2 [4 + 70]

∴ Using T

T

T

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

(T

⇒ a + 14d - a - 4d = - 30

⇒ 10d = - 30 ⇒ d = - 3

Substituting d = - 3 in (2), we get

a + 4d = 13

⇒ a + 4 (- 3) = 13

⇒ a + (- 12) = 13

⇒ a = 13 + 12 = 25

Now using S

S

= 21/2[50 + (- 60)]

Thus, the sum of the first fifteen terms = - 105.

Last term T

∵ Common difference (d) = 9.

∴ Using T

350 = 17 + (n - 1) × 9

⇒

⇒ n = 37 + 1 = 38

Thus, there are 38 terms.

Now, using, S_{n} = n/2 [a + l], we have

S_{38} = 38/2 [17 + 350]

= 19 [367] = 6973

Thus, the required sum of 38 terms

= 6973.**Q8. If the sum of the first 7 terms of an A.P. is 49 and that of the first 17 terms is 289, find the sum of n terms.****Sol. **Let the first term = a and the common difference = d.

∴ Using S_{n} = n/2 [2a + (n - 1) d]

∴ S_{7} = 7/2 [2a + 6d] = 49

⇒

⇒ 7 [a + 3d] = 49

⇒ a + 3d = 49/7 = 7

i.e., a + 3d = 7 ...(1)

Also S_{17} = 17/2 [2a + 16d] = 289

⇒

⇒ 17 [a + 8d] = 289

⇒ a + 8d = 289/17 = 17

⇒ a + 8d = 17 ...(2)

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

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

⇒ 5d = 10 ⇒ d = 2

From (1), we have

a + 3 (2) = 7

⇒ a + 6 = 7 ⇒ a = 7 - 6 = 1

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

Thus, the sum of n terms is n^{2}.**Q9. The first and last term of an A.P. is 4 and 81 respectively. If the common difference is 7, how many terms are there in the A.P. and what is their sum?****Sol. **Here, first term = 4 ⇒ a = 4 and d = 7.

Last term, l = 81 ⇒ T_{n} = 81

∵ T_{n} = a + (n - 1) d

∴ 81 = 4 + (n - 1) × 7

⇒ 81 - 4 = (n - 1) × 7

⇒ 77 = (n - 1) × 7

⇒

⇒ There are 12 terms.

Now, using

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

⇒ S_{12} = 12/2 (4 + 81)

⇒ S_{12} = 6 × 85 = 510

∴ The sum of 12 terms of the A.P. is 510.**Q10. The angles of a quadrilateral are in A.P. whose common difference is 15°. Find the angles.****Sol.** Let one of the angles = a

∵ The angles are in an A.P.

∴ The angles are:

a°, (a + d)°, (a + 2d)° and (a + 3d)°

∵ d = 15 [Given]

∴ The angles are:

a, (a + 15), [a + 2 (15)] and [a + 3 (15)]

i.e., a, (a + 15), (a + 30) and (a + 45).

∵ The sum of the angles of a quadrilateral is 360°.

∴ a + (a + 15) + (a + 30) + (a + 45) = 360°

⇒ 4a + 90° = 360°

⇒ 4a = 360° - 90° = 270°

⇒

∴ The four angles are:**Q11. The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.****Sol. **Let a, b, c are the angles of the triangle, such that

c = 2a ...(1)

Since a, b, c are in A.P.

Then ...(2)

From (1) and (2), we get

are the three angles of the triangle.

⇒2a + a + 2a + 4a = 360°

⇒ 9a = 360°

⇒ a = 360°/9 = 40°

∴ The smallest angle = 40°

The greatest angle = 2a = 2 × 40° = 80° The third angle

Thus the angles of the triangle are 40°, 60°, 80°.**Q12. Find the middle term of the A.P. 10, 7, 4, ....., - 62.****Sol.** Here, a = 10

d = 7 - 10 = - 3

T_{n} = (- 62)

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

- 62 = 10 + (n - 1) × (- 3)

⇒

⇒ n = 24 + 1 = 25

⇒ Number of terms = 25

∴ Middle term = term

Now T_{13} = 10 + 12d

= 10 + 12 (- 3)

= 10 - 36 = - 26

Thus, the middle term = - 26.**Q13. Find the sum of all three digit numbers which are divisible by 7.****Sol.** The three digit numbers which are divisible by 7 are:

105, 112, 119, ....., 994.

It is an A.P. such that

a = 105

d = 112 - 105 = 7

T_{n} = 994 = l

∵ T_{n} = a + (n - 1) × d

∴ 994 = 105 + (n - 1) × 7

⇒

⇒ n = 127 + 1 = 128

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

We have S_{128} = 128/2[105 + 994]

= 64 [1099]

= 70336

Thus, the required sum = 70336.**Q14. Find the sum of all the three digit numbers which are divisible by 9.****Sol. **All the three digit numbers divisible by 9 are:

117, 126, ....., 999 and they form an A.P.

Here, a = 108

d = 117 - 108 = 9

T_{n} = 999 = l

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

999 = 108 + (n - 1) (9)

⇒ 999 - 108 = (n - 1) × 9

⇒ 891 = (n - 1) × 9

⇒ n - 1 = 891/9 = 99

⇒ n = 99 + 1 = 100

Now, the sum of n term of an A.P. is given

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

∴ S_{100} = 100/2[108 + 999]

= 50 [1107]

= 55350

Thus, the required sum is 55350.**Q15. Find the sum of all the three digit numbers which are divisible by 11.****Sol.** All the three digit numbers divisible by 11 are 110, 121, 132, ....., 990.

Here, a = 110

d = 121 - 110 = 11

T_{n} = 990

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

990 = 110 + (n - 1) × 11

⇒

⇒ n = 80 + 1 = 81

Now, using S_{n} = n/2 [a + l], we have

S_{81} = 81/2 [110 + 990]

Thus, the required sum = 44550.**Q16. The sum of the first six terms of an AP is 42. The ratio of the 10th term to its 30th term is 1:3. Calculate the first term and 13th term of A.P.****Sol. ∵ **

∴ 6a + 15d = 4 2 ...(1)

Also, (a_{10}) : (a_{30}) = 1 : 3

or

⇒ 3(a + 9d) = a + 29d

⇒ 3a + 27d = a + 27d

⇒ 2a = 2 d

⇒ a = d ...(2)

From (1) 6d + 15d = 42 ⇒ d = 2

From (2) a = d ⇒ d = 2

Now, a_{13} = a + 12d

= 2 + 12 × 2 = 26

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