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**Q17. If S _{n }the sum of n terms of an A.P. is given by S_{n} = 3n^{2} - 4n, find the nth term.**

S

= 3 (n

= 3n

= 3n

∵ nth term = S

= 3n

= 3n

= 6n - 7.

Common difference be = d

∴ Using T

T

T

T

T

∵ T

∴ (a + 3d) + (a + 7d) = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12

[Dividing by 2] ...(1)

Also T

∴ (a + 5d) + (a + 9d) = 44

⇒ 2a + 14d = 44

⇒ a + 7d = 22

[Dividing by 2] ...(2)

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

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

⇒ 2d = 10 ⇒ d = 5

From (1), a + 5 (5) = 12

⇒ a = 12 - 25 = - 13

Since, the A.P. is given by:

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

∴ We have the required A.P. as:

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

or - 13, - 8, - 3, .....

Sol.

∴ S

= 5 (n

= 5n

= 5n

Now, nth term = S

∴ The required nth term

= [5n

= 10n - 2.

Now, using Tn = a + (n - 1) d, we have

T

T

T

T

∵ T

∴ a + 4d + a + 8d = 72

⇒ 2a + 12d = 72

⇒ a + 6d = 36

[Dividing by 2] ...(1)

Also T

⇒ 36 + a + 11d = 97 [From (1)]

⇒ a + 11d = 97 - 36

⇒ a + 11d = 61 ...(2)

Subtracting (1) from (2), we get

a + 11d - a - 6d = 61 - 36

⇒ 5d = 25

⇒ d = 25/5

From (1), we have

a + 11 (5) = 61

a + 55 = 61

⇒ a = 61 - 55 = 6

Now, a

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

∴ The required A.P. is:

6, (6 + 5), [6 + 2 (5)], [6 + 3 (5)], .....

or 6, 11, 16, 24, .....

And the common difference = d

∴

⇒ 10a + 45d = –150

⇒ 2a + 9d = –30 ...(1)

∵ The sum of next 10 terms

(i.e. S

⇒ 20a + 190d + 150 = –550

⇒ 2a + 19d + 15 = –55

⇒ 2a + 19d = – 55 – 15

⇒ 2a + 19d = –70 ...(2)

Subtracting (1) from (2), we get

From (1), 2(a) + 9(–4) = –30 or a = 6/2 = 3

Thus, AP is a, a + d, a + 2d ...

or 3, [3 + (–4)], [3 + 2(–4)], ...

or 3, –1, –5, ...

**Q22. Which term of the A.P. 3, 15, 27, 39, ..... will be 120 more than its 21st term?****Sol.** Let the 1st term is ‘a’ and common difference = d

∴ a = 3 and d = 15 - 3 = 12

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

∴ T_{21} = 3 + (21 - 1) × 12

= 3 + 20 × 12

= 3 + 240 = 243

Let the required term be the nth term.

∵ nth term = 120 + 21st term

= 120 + 243 = 363

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

⇒ 363 = 3 + (n - 1) × 12

⇒ 363 - 3 = (n - 1) × 12

⇒ n - 1 = 360/12 = 30

⇒ n = 30 + 1 = 31

Thus the required term is the 31st term of the A.P.**Q23. Which term of the A.P. 4, 12, 20, 28, ..... will be 120 more than its 21st term?****Sol.** Here, a = 4

d = 12 - 4 = 8

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

∴ T_{21} = 4 + (21 - 1) × 8

= 4 + 20 × 8 = 164

∵ The required nth term = T_{21} + 120

∴ nth term = 164 + 120 = 284

∴ 284 = a + (n - 1) d

⇒ 284 = 4 + (n - 1) × 8

⇒ 284 - 4 = (n - 1) × 8

⇒ n - 1 = 280/8 = 35

⇒ n = 35 + 1 = 36

Thus, the required term is the 36th term of the A.P.**Q24. The sum of n terms of an A.P. is 5n ^{2} - 3n. Find the A.P. Hence find its 10th term.**

S

∴ S

⇒ First term T1 = (a) = 2

S

= 20 - 6 = 14

⇒ Second term T

Now the common difference = T

⇒ d = 12 - 2 = 10

∵ An A.P. is given by

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

∴ The required A.P. is:

2, (2 + 10), [2 + 2 (10)], .....

⇒ 2, 12, 22, .....

Now, using T

T

= 2 + 9 × 10

= 2 + 90 = 92.

d = 10 - 8 = 2

T

Using T

⇒ 126 = 8 + (n - 1) × 2

⇒ n - 1 =

⇒ n = 59 + 1 = 60

∴ l = 60

Now 10th term from the end is given by

l - (10 - 1) = 60 - 9 = 51

Now, T

= 8 + 50 × 2

= 8 + 100 = 108

Thus, the 10th term from the end is 108.

S

∴ S

= 3 + 5 = 8

⇒ T

S

= 12 + 10 = 22

⇒ T

Now d = T

∵ An A.P. is given by,

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

∴ The required A.P. is:

8, (8 + 6), [8 + 2 (6)], .....

⇒ 8, 14, 20, .....

Now, using T

T

= 8 + 15 × 6 = 98

Thus, the 16th term of the A.P. is 98.

∴

Now, a

⇒

⇒

⇒

Using T

T

T

Since T

∴ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24 ⇒ a + 5d = 12 ...(1)

Also, T

∴ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44 ⇒ a + 7d = 22 ...(2)

Subtracting (2) from (1), we get,

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

⇒ 2d = 10 ⇒ d = 5

Now from (1),

a + 5 (5) = 12

⇒ a + 25 = 12 ⇒ a = - 13

∴ First term (T

Second term (T

= - 13 + 5 = - 8

Third term T

= - 13 + 10 = - 3

First term T

nth term T

∵ S

∴ Using, S

S

⇒

⇒

Now, T

⇒ a + 5d = 33

⇒ 8 + 5d = 33

⇒ 5d = 33 - 8 = 25

⇒ d = 25/5 = 5

Thus, n = 6 and d = 5.

a = 63

d = 65 - 63 = 2

∴ T

⇒ T

For the 2nd A.P.

a = 3

d = 10 - 3 = 7

∴ T

⇒ T

∵ [T

∴ 63 + (n - 1) × 2 = 3 + (n - 1) × 7

⇒ 63 - 3 + (n - 1) × 2 = (n - 1) 7

⇒ 60 + (n - 1) × 2 - (n - 1) × 7 = 0

⇒ 60 + (n - 1) [2 - 7] = 0

⇒ 60 + (n - 1) × (- 5) = 0

⇒ (n - 1) = -60/-5 = 12

⇒ n = 12 + 1 = 13

Thus, the required value of n is 13.

∴ nth term = a + (n - 1) d

And mth term = a + (m - 1) d

Also,

(m + n)th term = a + (m + n - 1) d ...(1)

∵ m (mth term) = n (nth term)

∴ m [a + (m - 1) d] = n [a + (n - 1) d]

⇒ ma + m (m - 1) d = na + n (n - 1) d

⇒ ma + (m

⇒ ma - na + (m

⇒ a [m - n] + [m

⇒ a [m - n] + [(m

⇒ a [m - n] + [(m + n) (m - n) - (m - n)] d = 0

⇒ a [m - n] + (m - n) [m + n - 1] d = 0

Dividing throughout by (m - n), we have:

a + [m + n - 1] d = 0

⇒ a + [(m + n) - 1] d = 0 ...(2)

⇒ (m + n) th term = 0 [From (1) and (2)]

And the nth term = - 17 = l

Using T

- 17 = 25 + (n - 1) d

⇒ (n - 1) d = - 17 - 25 = - 42

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

⇒ 60 = n/2 [25 + (- 17)]

⇒

⇒ 60 = 4n ⇒ n = 60/4 = 15

From (1), we have

Thus, n = 15 and d = - 3**Q33. In an A.P., the first term is 22, nth term is - 11 and sum of first n terms is 66. Find n and d, the common difference. ****Sol.** We have

1st term (T_{1}) = 22 ⇒ a = 22

Last term (T_{n}) = - 11 ⇒ l = - 11

Using, S_{n} = n/2 [a + l], we have:

66 = n/2 [22 + (- 11)]

⇒ 66 × 2 = n [11]

⇒

Again using

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

We have:

T_{12 }= 22 + (12 - 1) × d

- 11 = 22 + 11d [∵ nth term = - 11]

⇒ 11d = - 22 - 11 = - 33

⇒ d = -33/11 = -3

Thus, n = 12 and d = - 3

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