Question 1: Find:

i) {10}^{th} term of A.P 1, 4, 7, 10, \ldots           ii) {18}^{th} term of A.P \sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, \ldots

iii) n^{th} term of A.P 13, 8, 3, -2, \ldots

Answer:

i)       Given A.P. series 1, 4, 7, 10, \ldots

Therefore  a = 1

Common difference (d) = 4-1 = 3

We know, a_n = a + (n-1) d

\therefore a_{10} = 1+ ( 10 - 1) (3) = 1 + 27 = 28 

ii)      Given A.P. series \sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, \ldots

Therefore  a = \sqrt{2}

Common difference (d) = 3\sqrt{2} - \sqrt{2} = 2\sqrt{2}

We know, a_n = a + (n-1) d

\therefore a_{18} = \sqrt{2} + ( 18 - 1) (2\sqrt{2}) = \sqrt{2} + 34\sqrt{2} = 35\sqrt{2} 

iii)    Given A.P. series 13, 8, 3, -2, \ldots

Therefore  a = 13

Common difference (d) = 8-13 = -5

We know, a_n = a + (n-1) d

\therefore a_{n} = 13+ ( n - 1) (-5) = -5n+18 

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Question 2: In an A.P., show that a_{m + n} + a_{m - n} = 2a_m .

Answer:

Let the first term be a and the common difference be d

LHS = a_{m + n} + a_{m - n}

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

= 2a + ( m+n-1+m-n-1)d

= 2a + 2 (m-1) d

= 2 [ a+ ( m-1)d ]

= 2 a_m . Hence proved.

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Question 3:

i) Which term of the A.P. 3, 8, 13, \ldots is 248 ?

ii) Which term of the A.P. 84, 80,76, \ldots is 0 ?

iii) Which term of the A.P. 4, 9,14, \ldots is 254 ?

Answer:

i)       Given A.P. series 3, 8, 13, \ldots is 248 ?

\Rightarrow a = 3            d = (8-3) = 5           a_n = 248

We know a_n = a + (n-1) d

\Rightarrow 248 = 3 + (n-1)(5)

\Rightarrow 49 = n - 1

\Rightarrow n = 50

Hence 248 is the 50^{th} term.

ii)      Given A.P. series 84, 80,76, \ldots is 0 ?

\Rightarrow a = 84            d = (80 - 84) = -4           a_n = 0

We know a_n = a + (n-1) d

\Rightarrow 0 = 84 + (n-1)(-4)

\Rightarrow 21 = n - 1

\Rightarrow n = 22

Hence 0 is the 22^{nd} term.

iii)     Given A.P. series 4, 9,14, \ldots is 254 ?

\Rightarrow a = 4            d = (9-4) = 5           a_n = 254

We know a_n = a + (n-1) d

\Rightarrow 254 = 4 + (n-1)(5)

\Rightarrow 50 = n - 1

\Rightarrow n = 51

Hence 254 is the 51^{st} term.

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Question 4:

(i) Is 58 a term of the A.P. 7,10,13, \ldots ?

(ii) Is 302 a term of the A.P. 3,8,13, \ldots ?

Answer:

(i)      Given A.P. series 7,10,13, \ldots ?

\Rightarrow a = 7            d = (10-7) = 3           a_n = 68

We know a_n = a + (n-1) d

\Rightarrow 68 = 7 + (n-1)(3)

\Rightarrow \frac{61}{3} = n - 1

\Rightarrow n = \frac{61}{3} + 1

Since n is not a natural number, 68 is NOT a term in the given A.P.

(ii)    Given A.P. series 3,8,13, \ldots ?

\Rightarrow a = 3            d = (8-3) = 5           a_n = 302

We know a_n = a + (n-1) d

\Rightarrow 302 = 3 + (n-1)(5)

\Rightarrow \frac{299}{5} = n - 1

\Rightarrow n = \frac{299}{5} + 1

Since n is not a natural number, 302 is NOT a term in the given A.P.

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Question 5:

i) Which term of the sequence 24, 23 \frac{1}{4} , 22 \frac{1}{2} , 21 \frac{3}{4} , \ldots   is the first negative term?

ii) Which term of  the sequence 12+8i, 11,+6i, 10+4i , \ldots   is (a) purely real (b) purely imaginary ?

Answer:

i)      Given series 24, 23 \frac{1}{4} , 22 \frac{1}{2} , 21 \frac{3}{4} , \ldots

\Rightarrow a = 24            d = (23 \frac{1}{4} -24) = - \frac{3}{4}

Let the first negative term is a_n .

Therefore a_n  < 0

\Rightarrow a + (n-1) d < 0

\Rightarrow 24 + (n-1) ( - \frac{3}{4} ) < 0

\Rightarrow 24 + \frac{3}{4} < \frac{3n}{4}

\Rightarrow \frac{99}{4} < \frac{3n}{4}

\Rightarrow 99 < 3n

Therefore n > 33

Thus the 34^{th} term is the first negative term of the given AP.

ii)      Given series 12+8i, 11,+6i, 10+4i , \ldots

\Rightarrow a = 12 + 8i

d = ( 11+6i) - ( 12 + 8i) = -1 -2i

Let the real term be a_n

a_n = a + (n-1)d

= ( 12 + 8i) + ( n-1) ( -1 -2i)

= 12 + 8i + 1 + 2i - n - 2in

= ( 12 + 1 - n) + (10i - 2in)

= (13-n) + i ( 10-2n)

a) for a_n to be real, 10-2n = 0 \Rightarrow n = 5

b) for a_n to be imaginary, 13-n = 0 \Rightarrow n = 13

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Question 6:

i) How many terms are there in the A.P. 7, 10, 13, \ldots ,  43 ?

ii) How many terms are there in the A.P. -1, \frac{-5}{6} , \frac{-2}{3} , \frac{-1}{2} , \ldots ,  \frac{10}{3} ?

Answer:

i)       Given A.P. series 7, 10, 13, \ldots ,  43

Therefore a = 7      d= ( 10 - 7) = 3        a_n = 43

We know a_n = a + (n-1)d

\Rightarrow 43 = 7 + (n-1) (3)

\Rightarrow 36 = (n-1) (3)

\Rightarrow n -1 = 12

\Rightarrow n = 13

Therefore there are 13 terms in the given A.P.

ii)     Given A.P. series -1, \frac{-5}{6} , \frac{-2}{3} , \frac{-1}{2} , \ldots ,  \frac{10}{3} ?

Therefore a = -1      d= \Big[ \frac{-5}{6} - (-1) \Big]  = \frac{1}{6}         a_n = \frac{10}{3}

We know a_n = a + (n-1)d

\Rightarrow \frac{10}{3} = -1 + (n-1) ( \frac{1}{6} )

\Rightarrow \frac{13}{3} = (n-1) ( \frac{1}{6} )

\Rightarrow n -1 = 26

\Rightarrow n = 27

Therefore there are 27 terms in the given A.P.

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Question 7: The first term of an A.P. is 5 , the common difference is 3 and the last term is 80 ; find the number of terms.

Answer:

Given: a = 5          d = 3           a_n = 80

We know a_n = a + (n-1)d

\Rightarrow 80 = 5 + ( n-1) (3)

\Rightarrow (n-1) (3) = 75

\Rightarrow n-1 = 25

\Rightarrow n = 26

Therefore there are 26 terms in the given A.P.

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Question 8: The 6^{th} and 17^{th} terms of an A.P. are 19 and 41 respectively, find the 40^{th} term.

Answer:

Given a_6 = 19    and a_{17} = 41

We know a_n = a + (n-1)d

\Rightarrow a + ( 6-1) d = 19

\Rightarrow a + 5d = 19      … … … … … i)

\Rightarrow a + ( 17-1) d = 41

\Rightarrow a + 16d = 41      … … … … … ii)

Solving i) and ii)

19-5d = 41 - 16d

\Rightarrow 11d = 22

\Rightarrow d = 2

Therefore a = 19 - 5( 2) = 9

Hence 40^{th} is a_{40} = 9 + (40-1) (2) = 87

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Question 9: If 9^{th} term of an A.P. is zero, prove that its 29^{th} term is double the 19^{th} term.

Answer:

Given: a_9 = 0

\therefore a + ( 9-1) d = 0

\Rightarrow a + 8d = 0

\Rightarrow a = -8d

To prove: a _{29} = 2 a_{19}

\Rightarrow a + ( 29-1) d = 2 [ a + (19-1) d]

\Rightarrow a + 28d = 2a + 36d

Substituting a = -8d

\Rightarrow -8d + 28d  = 2 (-8d) + 36d

\Rightarrow 20 d = 20 d .    Hence proved.

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Question 10: If 10 times the 10^{th} term of an A.P. is equal to 15 times the 15^{th} term, show that 25^{th} term of the A.P. is zero.

Answer:

Given 10a_{10} = 15 a_{15}

We know a_n = a + (n-1)d

\Rightarrow 10 [ a+ ( 10 - 1) d] = 15 [ a+ ( 15 - 1) d]

\Rightarrow 10 a + 90d = 15a + 210 d

\Rightarrow 5a + 120 d = 0

\Rightarrow a + 24 d = 0

\Rightarrow a = - 24 d

Now a_{25} = a + ( 25 - 1) d = -24d + 24 d = 0 . Hence proved.

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Question 11: The 10^{th} and 18^{th} terms of an A.P. are 41 and 73 respectively. Find 26^{th} term.

Answer:

Given a_{10} = 41 and a_{18} = 73

\therefore a + ( 10 -1) D = 41

\Rightarrow a + 9d = 41      … … … … … i)

Also a + ( 18 - 1) d = 73

\Rightarrow a + 17d = 73      … … … … … ii)

Solving i) and ii) we get

41- 9d = 73 - 17d

\Rightarrow 8d = 32

\Rightarrow d = 4

Substituting from i) we get

a+ 9(4) = 41 \Rightarrow a = 5

Therefore a_{26} = 5 + ( 26-1) (4)  = 105

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Question 12: In a certain A.P. the 24^{th} term is twice the 10^{th} term. Prove that the 72^{nd} term is twice the 34^{th} term.

Answer:

Given a_{24} = 2a_{10}

\Rightarrow a + (24-1) d = 2 [ a + (10-1) d]

\Rightarrow a + 23 d = 2a + 18 d

\Rightarrow a = 5d      … … … … … i)

To prove: a_{72} = 2 a_{34}

\Rightarrow a + ( 72 -1) d = 2 [ a + ( 34 - 1) d]

\Rightarrow a + 71 d = 2a + 66d

Substituting from i) we get

\Rightarrow 5d + 71 d = 2 ( 5d) + 66d

\Rightarrow 76d = 76 d . Hence proved.

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Question 13: lf (m + 1)^{th} term of an A.P. is twice the (n + 1)^{th} term, prove that (3m+ 1)^{th} term is twice the (m+n+1)^{th} term.

Answer:

Given a_{m+1} = 2a_{n+1}

\Rightarrow [ a + ( m+1-1) d ] = 2 [ a + (n+1-1)d]

\Rightarrow a + md = 2a + 2nd

\Rightarrow a = (m-2n) d      … … … … … i)

To prove a_{3m+1} = 2 a_{m+n+1}

\Rightarrow a + (3m+1-1)d = 2 [ a + (m+n+1-1) d

\Rightarrow a + 3md = 2a + 2(m+n) d

Substituting from i)

\Rightarrow (m-2n) d + 3md = 2 [(m-2n)d] + 2(m-n)d

\Rightarrow 4md - 2nd = 2md - 4nd + 2md + 2nd

\Rightarrow 4md - 2nd = 4md - 2nd . Hence proved.

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Question 14: If the n^{th} term of the A.P. 9, 7, 5, \ldots is same as the n^{th} term of the A.P. 15, 12, 9, \ldots find n .

Answer:

n^{th} term of sequence 9, 7, 5, \ldots

\Rightarrow a = 9, d = (7-9) = -2

\therefore a_n = a + (n-1) d

= 9 + ( n-1) ( -2)

= 11 - 2n

n^{th} term of sequence 15, 12, 9, \ldots

\Rightarrow a = 15, d = (12-15) = -3

\therefore a_n = a + (n-1) d

= 15 + ( n-1) ( -3)

= 18 - 3n

Given: 11 - 2n = 18-3n \Rightarrow n = 7

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Question 15: Find the 12^{th} term from the end of the following arithmetic progressions: (i) 3,5,7,9, \ldots , 201 (ii) 3,8,13 , \ldots , 253 (iii) 1,4,7,10, \ldots ,88

Answer:

(i)      Given series 3,5,7,9, \ldots , 201

a = 3         d = 5 - 3 = 2           l = 201

n^{th} term from the end = l - ( n-1) d

Therefore 12^{th} term from the end = 201 - ( 12-1)(2) = 201 - 22 = 179

(ii)     Given series 3, 8,13 , \ldots , 253

a = 3         d = 8 - 3 = 5           l = 253

n^{th} term from the end = l - ( n-1) d

Therefore 12^{th} term from the end = 252 - ( 12-1)(5) = 253 - 55 = 198

(iii)    Given series 1,4,7,10, \ldots ,88

a = 1         d = 4 - 1 = 3           l = 88

n^{th} term from the end = l - ( n-1) d

Therefore 12^{th} term from the end = 88 - ( 12-1)(3) = 88 - 33 = 55

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Question 16: The 4^{th} term of an A.P. is three times the first and the 7^{th} term exceeds twice the third term by 1 . Find the first term and the common difference.

Answer:

Given a_4 = 3a

\Rightarrow a+ ( 4 - 1) d = 3a

\Rightarrow 3d = 2a       … … … … … i)

Also a_7 - 2a_3 = 1

\Rightarrow a+(7-1)d - 2 [ a + ( 3-1) d ] = 1

\Rightarrow a + 6d - 2 ( a + 2d) = 1

\Rightarrow a + 6d - 2a - 4d = 1

\Rightarrow 2d - a = 1       … … … … … ii)

Substituting in i) we get

3d = 2 ( 2d -1 )

\Rightarrow 3d = 4d - 2

\Rightarrow d = 2

\therefore a = 2(2) - 1 = 4 -1 = 3

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Question 17: Find the second term and n^{th} term of an A.P. whose 6^{th} term is 12 and the 8^{th} term is 22 .

Answer:

We know a_n = a + (n-1)d

Given, a_6 = 12

\Rightarrow a + ( 6-1) d = 12

\Rightarrow a + 5d = 12       … … … … … i)

Also a_8 = 22

\Rightarrow a + ( 8-1) d = 22

\Rightarrow a + 7d = 22       … … … … … ii)

Solving i) and ii) we get

12-5d = 22-7d

2d = 10

d = 5

\therefore a = 12 - 5 ( 5) = 12 - 25 = - 13

Hence a_2 = -13 + ( 2-1) (5) = -13 + 5 = -8

a_n = -13 + ( n-1)(5) = -13 + 5n - 5 = -18 + 5n

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Question 18: How many numbers of two digit are divisible by 3 ?

Answer:

The sequence would be 12, 15, 18, \ldots , 96, 99

Therefore a = 12 , \ \ \ \ d = ( 15-12) = 3 \ \ \ \ \ a_n = 99

We know a_n = a + (n-1)d

\therefore 99 = 12 + (n-1) (3)

\Rightarrow 90 = 3n

\Rightarrow n = 30

Therefore there are 30 such numbers which are two digit are divisible by 3

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Question 19: An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32^{nd} term.

Answer:

Given a = 7, \ \ \ \ n = 60 , \ \ \ \ \ l = 125

We know l = a + ( n-1) d

\Rightarrow 125 = 7 + ( 60-1) d

\Rightarrow 118 = 59 d

\Rightarrow d = 2

We know a_n = a + (n-1)d

\therefore a_{32} = 7 + ( 32-1) (2) = 69

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Question 20: The sum of 4^{th}  and 8^{th} terms of an A.P. is 24 and the sum of the 6^{th} and 10^{th} terms is 34 . Find the first term and the common difference of the A.P.

Answer:

Given a_4 + a_8 = 24

\Rightarrow a + ( 4-1) d + a + ( 8-1) d = 24

\Rightarrow 2a + 10d = 24

\Rightarrow a + 5d = 12

Also a_6 + a_{10} = 34

\Rightarrow a + ( 6-1) d + a + ( 10-1) d = 34

\Rightarrow 2a + 14d = 34

\Rightarrow a + 7d = 17

Solving i) and ii) we get

12-5d = 17 - 7d

\Rightarrow 2d = 5

\Rightarrow d = \frac{5}{2}

Substituting in i) we get

a = 2 - 5 ( \frac{5}{2} ) = 12 - \frac{25}{2} = \frac{-1}{2}

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Question 21: How many numbers are there between 1 and 1000 which when divided by 7 leave remainder 4 ?

Answer:

If a number N is divided by 7 and leaves 4 as remainder, it can be represented as N = 7q + 4 , \ \ q \in N

Therefore the sequence would be 4, 11, 18, \ldots, 998

Hence a = 4, \ \ \ \ d = ( 11- 4) = 7, \ \ \ \ \ l = 998

We know l = a + ( n-1) d

\Rightarrow 998 = 4 + ( n-1) (7)

\Rightarrow 1001 = 7n

\Rightarrow n = 143

Hence there are 143 numbers between 1 and 1000 which when divided by 7 leave remainder 4

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Question 22: The first and the last terms of an A.P. are a and l respectively. Show that the sum of n^{th} term from the beginning and n^{th} term from the end is a + l .

Answer:

Given: first term = a and last term = l

a_n = a + (n-1) d

n^{th} term from end = l - (n-1)d

Therefore the sum of the two terms = a + (n-1) d + l - (n-1)d = a+l

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Question 23: If an A.P. is such that \frac{a_4}{a_7} = \frac{2}{3} , find \frac{a_6}{a_8} .

Answer:

Given \frac{a_4}{a_7} = \frac{2}{3}

\Rightarrow \frac{a + (4-1)d}{a + (7-1) d} = \frac{2}{3}

\Rightarrow 3(a+3d) = 2(9+6d)

\Rightarrow 3a + 9d = 2a +12d

\Rightarrow a = 3d

Therefore \frac{a_6}{a_8} = \frac{a+(6-1)d}{a+(8-1)d}   = \frac{a+5d}{a+7d} = \frac{3d+5d}{3d+7d} = \frac{8}{10} = \frac{4}{5}

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Question 24: If  \theta_1,  \theta_2,  \theta_3, \ldots \theta_n are in AP, whose common difference is d , show that, \sec \theta_1 \ \sec \theta_2 + \sec \theta_2 \ \sec \theta_3 + \ldots + \sec \theta_{n-1} \ \sec \theta_n = \frac{\tan \theta_n - \tan \theta_1}{\sin d}

Answer:

Given \theta_1,  \theta_2,  \theta_3, \ldots \theta_n are in AP

\therefore d = \theta_2 -  \theta_1= \theta_3 -  \theta_2=  \ldots = \theta_{n} -  \theta_{n-1}

LHS = \sec \theta_1 \ \sec \theta_2 + \sec \theta_2 \ \sec \theta_3 + \ldots + \sec \theta_{n-1} \ \sec \theta_n 

= \frac{1}{\cos \theta_1 \cdot \cos \theta_2} + \frac{1}{\cos \theta_2 \cdot \cos \theta_3} + \ldots + \frac{1}{\cos \theta_{n-1} \cdot \cos \theta_n}

= \frac{1}{\sin d} \Big[  \frac{\sin d}{\cos \theta_1 \cdot \cos \theta_2} + \frac{\sin d}{\cos \theta_2 \cdot \cos \theta_3} + \ldots + \frac{\sin d}{\cos \theta_{n-1} \cdot \cos \theta_n}  \Big]

= \frac{1}{\sin d} \Big[  \frac{\sin (\theta_2 -  \theta_1)}{\cos \theta_1 \cdot \cos \theta_2} + \frac{\sin (\theta_3 -  \theta_2)}{\cos \theta_2 \cdot \cos \theta_3} + \ldots + \frac{\sin (\theta_n -  \theta_{n-1})}{\cos \theta_{n-1} \cdot \cos \theta_n}  \Big]

= \frac{1}{\sin d} \Big[  \frac{\sin \theta_2 \cdot \cos \theta_1 -  \cos \theta_2 \cdot \sin \theta_1 }{\cos \theta_1 \cdot \cos \theta_2} + \frac{\sin \theta_3 \cdot \cos \theta_2 -  \cos \theta_3 \cdot \sin \theta_2}{\cos \theta_2 \cdot \cos \theta_3} + \ldots + \frac{\sin \theta_n \cdot \cos \theta_{n-1} -  \cos \theta_n \cdot \sin \theta_{n-1}}{\cos \theta_{n-1} \cdot \cos \theta_n}  \Big]

= \frac{1}{\sin d} [   \tan \theta_2 - \tan \theta_1 + \tan \theta_3 - \tan \theta_2 + \ldots + \tan \theta_n - \tan \theta_{n-1} ]

= \frac{1}{\sin d} [ \tan \theta_n - \tan \theta_1 ]

= \frac{  \tan \theta_n - \tan \theta_1}{\sin d} = RHS. Hence proved.