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**Short Answer Type Questions****Note :** For an A.P. with the 1st term and common difference â€˜aâ€™ and â€˜dâ€™ respectively, we have :

(a) n^{th} term from the end = (m â€“ n +1)th term from the beginning, where m is the number of terms in the A.P.

â‡’ n^{th} term from the end = (a) + (m â€“ n)d

(b) If â€˜lâ€™ is the last term of the A.P., then

n^{th} term from the end is the nth term of an A.P. whose first term is â€˜lâ€™ and common difference is â€˜â€“dâ€™

â‡’ nth term from the end = l + (n â€“ 1) (â€“d)**Q1. If 9th term of an A.P. is zero, prove that its 29th term is double of its 19th term****Sol.** Let â€˜aâ€™ be the first term and â€˜dâ€™ be the common difference.

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

T_{9} = a + 8d â‡’ a + 8d = 0 ...(1) [âˆµ T_{9} = 0 Given]

T_{19} = a + 18d = (a + 8d) + 10d = (0) + 10d = 10d ...(2)

[âˆµ a + 8d = 0]

T_{29} = a + 28d

= (a + 8d) + 20d

= 0 + 20d = 20d [âˆµ a + 8d = 0]

= 2 Ã— (10d) = 2 (T19) [âˆµ T_{19} = 10d]

â‡’ T_{29} = 2 (T_{19})

Thus, the 29th term of the A.P. is double of its 19th term.**Q2. If T _{n} = 3 + 4n then find the A.P. and hence find the sum of its first 15 terms.**

âˆµ T

âˆ´ T

T

But it is given that

T

âˆ´ T

â‡’ First term, a = 7

Also, T

âˆ´d = T

Now, using S

we get

S

Thus, the sum of first 15 terms = 525.

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

âˆ´ a = 3

d = 15 - 3 = 12

âˆ´ Using, T

T

= 3 + (52 Ã— 12)

= 3 + 624 = 627

Now, T

Let the required term be T

âˆ´ T

or a + (n - 1) d = 747

âˆ´ 3 + (n - 1) Ã— 12 = 747

â‡’ (n - 1) Ã— 12 = 747 - 3 = 744

â‡’ n - 1 = 744/12 = 62

â‡’ n = 62 + 1 = 63

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

Using T

T

â‡’ 31 = a + 9d ...(1)

Also T

â‡’ 66 = a + 14d ...(2)

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

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

â‡’ 5d = 35

â‡’ d = 35/5 = 7

âˆ´ From (1), a + 9d = 31

â‡’ a + 9 (7) = 31

â‡’ a + 63 = 31

â‡’ a = 31 - 63

â‡’ 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

âˆ´ Using T

âˆ´ T

â‡’ 37 = a + 7d ...(1)

Also T

And T

According to the question,

T

â‡’ a + 14d = a + 11d + 15

â‡’ a - a + 14d - 11d = 15

â‡’ 3d = 15 â‡’ d = 15/3 = 5

From (1), we have:

a + 7 (5) = 37

â‡’ a + 35 = 37

â‡’ a = 37 - 35 = 2

Since an A.P. is given by :

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

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

Now, using S

âˆ´ S

= 15/2 [4 + 70]

âˆ´ Using T

T

T

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

(T

â‡’ a + 14d - a - 4d = - 30

â‡’ 10d = - 30 â‡’ d = - 3

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

a + 4d = 13

â‡’ a + 4 (- 3) = 13

â‡’ a + (- 12) = 13

â‡’ a = 13 + 12 = 25

Now using S

S

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

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

Last term T

âˆµ Common difference (d) = 9.

âˆ´ Using T

350 = 17 + (n - 1) Ã— 9

â‡’

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

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

âˆ´ S_{7} = 7/2 [2a + 6d] = 49

â‡’

â‡’ 7 [a + 3d] = 49

â‡’ a + 3d = 49/7 = 7

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

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

â‡’

â‡’ 17 [a + 8d] = 289

â‡’ a + 8d = 289/17 = 17

â‡’ a + 8d = 17 ...(2)

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

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

â‡’ 5d = 10 â‡’ d = 2

From (1), we have

a + 3 (2) = 7

â‡’ a + 6 = 7 â‡’ 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. are 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 â‡’ a = 4 and d = 7.

Last term, l = 81 â‡’ T_{n} = 81

âˆµ T_{n} = a + (n - 1) d

âˆ´ 81 = 4 + (n - 1) Ã— 7

â‡’ 81 - 4 = (n - 1) Ã— 7

â‡’ 77 = (n - 1) Ã— 7

â‡’

â‡’ There are 12 terms.

Now, using

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

â‡’ S_{12} = 12/2 (4 + 81)

â‡’ S_{12} = 6 Ã— 85 = 510

âˆ´ 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 15Â°. Find the angles.****Sol.** Let one of the angles = a

âˆµ The angles are in an A.P.

âˆ´ The angles are:

aÂ°, (a + d)Â°, (a + 2d)Â° and (a + 3d)Â°

âˆµ d = 15 [Given]

âˆ´ The angles are:

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

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

âˆµ The sum of the angles of a quadrilateral is 360Â°.

âˆ´ a + (a + 15) + (a + 30) + (a + 45) = 360Â°

â‡’ 4a + 90Â° = 360Â°

â‡’ 4a = 360Â° - 90Â° = 270Â°

â‡’

âˆ´ 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.

â‡’2a + a + 2a + 4a = 360Â°

â‡’ 9a = 360Â°

â‡’ a = 360Â°/9 = 40Â°

âˆ´ The smallest angle = 40Â°

The greatest angle = 2a = 2 Ã— 40Â° = 80Â° The third angle

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

d = 7 - 10 = - 3

T_{n} = (- 62)

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

- 62 = 10 + (n - 1) Ã— (- 3)

â‡’

â‡’ n = 24 + 1 = 25

â‡’ Number of terms = 25

âˆ´ 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

âˆµ T_{n} = a + (n - 1) Ã— d

âˆ´ 994 = 105 + (n - 1) Ã— 7

â‡’

â‡’ 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)

â‡’ 999 - 108 = (n - 1) Ã— 9

â‡’ 891 = (n - 1) Ã— 9

â‡’ n - 1 = 891/9 = 99

â‡’ n = 99 + 1 = 100

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

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

âˆ´ 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

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

990 = 110 + (n - 1) Ã— 11

â‡’

â‡’ 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 first six terms of an AP is 42. The ratio of 10th term to its 30th term is 1 : 3. Calculate the first term and 13th term of A.P.****Sol. âˆµ **

âˆ´ 6a + 15d = 4 2 ...(1)

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

or

â‡’ 3(a + 9d) = a + 29d

â‡’ 3a + 27d = a + 27d

â‡’ 2a = 2 d

â‡’ a = d ...(2)

From (1) 6d + 15d = 42 â‡’ d = 2

From (2) a = d â‡’ d = 2

Now, a_{13} = a + 12d

= 2 + 12 Ã— 2 = 26

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