RD Chapter 19- Arithmetic Progressions Ex-19.1 |
RD Chapter 19- Arithmetic Progressions Ex-19.3 |
RD Chapter 19- Arithmetic Progressions Ex-19.4 |
RD Chapter 19- Arithmetic Progressions Ex-19.5 |
RD Chapter 19- Arithmetic Progressions Ex-19.6 |
RD Chapter 19- Arithmetic Progressions Ex-19.7 |

Find:

(i) 10^{th} term of the A.P. 1, 4, 7, 10, …..

(ii) 18^{th} term of the A.P. √2, 3√2, 5√2, …

(iii) nth term of the A.P 13, 8, 3, -2, ….

**Answer
1** :

(i) 10^{th} termof the A.P. 1, 4, 7, 10, …..

Arithmetic Progression(AP) whose common difference is = a_{n} – a_{n-1} wheren > 0

Let us consider, a = a_{1} =1, a_{2} = 4 …

So, Common difference,d = a_{2} – a_{1} = 4 – 1 = 3

To find the 10^{th} termof A.P, firstly find a_{n}

By using the formula,

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

= 1 + (n-1) 3

= 1 + 3n – 3

= 3n – 2

When n = 10:

a_{10} =3(10) – 2

= 30 – 2

= 28

Hence, 10^{th} termis 28.

(ii) 18^{th} termof the A.P. √2, 3√2, 5√2, …

Arithmetic Progression(AP) whose common difference is = a_{n} – a_{n-1} wheren > 0

Let us consider, a = a_{1} =√2, a_{2} = 3√2 …

So, Common difference,d = a_{2} – a_{1} = 3√2 – √2 = 2√2

To find the 18^{th} termof A.P, firstly find a_{n}

By using the formula,

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

= √2 + (n – 1) 2√2

= √2 + 2√2n – 2√2

= 2√2n – √2

When n = 18:

a_{18} =2√2(18) – √2

= 36√2 – √2

= 35√2

Hence, 10^{th} termis 35√2

(iii) nth term of the A.P13, 8, 3, -2, ….

Arithmetic Progression(AP) whose common difference is = a_{n} – a_{n-1} wheren > 0

Let us consider, a = a_{1} =13, a_{2} = 8 …

So, Common difference,d = a_{2} – a_{1} = 8 – 13 = -5

To find the n^{th} termof A.P, firstly find a_{n}

By using the formula,

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

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

= 13 – 5n + 5

= 18 – 5n

Hence, n^{th} termis 18 – 5n

In an A.P., show that a_{m+n} + a_{m–n} = 2a_{m}.

**Answer
2** :

We know the first termis ‘a’ and the common difference of an A.P is d.

Given:

a_{m+n} +a_{m–n} = 2a_{m}

By using the formula,

a_{n} = a+ (n – 1)d

Now, let us take LHS:a_{m+n} + a_{m-n}

a_{m+n} +a_{m-n} = a + (m + n – 1)d + a + (m – n – 1)d

= a + md + nd – d + a+ md – nd – d

= 2a + 2md – 2d

= 2(a + md – d)

= 2[a + d(m – 1)] {∵ a_{n} = a + (n – 1)d}

a_{m+n} +a_{m-n} = 2a_{m}

Hence Proved.

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

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

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

**Answer
3** :

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

Given A.P is 3, 8,13,…

Here, a_{1} =a = 3, a_{2} = 8

Common difference, d =a_{2} – a_{1} = 8 – 3 = 5

We know, a_{n} =a + (n – 1)d

a_{n} = 3+ (n – 1)5

= 3 + 5n – 5

= 5n – 2

Now, to find whichterm of A.P is 248

Put a_{n} =248

∴ 5n – 2 = 248

= 248 + 2

= 250

= 250/5

= 50

Hence, 50^{th} termof given A.P is 248.

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

Given A.P is 84, 80,76,…

Here, a_{1} =a = 84, a_{2} = 88

Common difference, d =a_{2} – a_{1} = 80 – 84 = -4

We know, a_{n} =a + (n – 1)d

a_{n} =84 + (n – 1)-4

= 84 – 4n + 4

= 88 – 4n

Now, to find whichterm of A.P is 0

Put a_{n} =0

88 – 4n = 0

-4n = -88

n = 88/4

= 22

Hence, 22^{nd} termof given A.P is 0.

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

Given A.P is 4, 9,14,…

Here, a_{1} =a = 4, a_{2} = 9

Common difference, d =a_{2} – a_{1} = 9 – 4 = 5

We know, a_{n} =a + (n – 1)d

a_{n} = 4+ (n – 1)5

= 4 + 5n – 5

= 5n – 1

Now, to find whichterm of A.P is 254

Put a_{n} =254

5n – 1 = 254

5n = 254 + 1

5n = 255

n = 255/5

= 51

Hence, 51^{st} termof given A.P is 254.

(i) Is 68 a term of the A.P. 7, 10, 13,…?

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

**Answer
4** :

(i) Is 68 a term of theA.P. 7, 10, 13,…?

Given A.P is 7, 10,13,…

Here, a_{1} =a = 7, a_{2} = 10

Common difference, d =a_{2} – a_{1} = 10 – 7 = 3

We know, a_{n} =a + (n – 1)d [where, a is first term or a_{1} and d is commondifference and n is any natural number]

a_{n} = 7+ (n – 1)3

= 7 + 3n – 3

= 3n + 4

Now, to find whether68 is a term of this A.P. or not

Put a_{n} =68

3n + 4 = 68

3n = 68 – 4

3n = 64

n = 64/3

64/3 is not anatural number

Hence, 68 is not aterm of given A.P.

(ii) Is 302 a term of theA.P. 3, 8, 13,…?

Given A.P is 3, 8,13,…

Here, a_{1} =a = 3, a_{2} = 8

Common difference, d =a_{2} – a_{1} = 8 – 3 = 5

We know, a_{n} =a + (n – 1)d

a_{n} = 3+ (n – 1)5

= 3 + 5n – 5

= 5n – 2

To find whether 302 isa term of this A.P. or not

Put a_{n} =302

5n – 2 = 302

5n = 302 + 2

5n = 304

n = 304/5

304/5 is not anatural number

Hence, 304 is not aterm of given A.P.

(i) Which term of the sequence 24, 23 ¼, 22 ½, 21 ¾ is the firstnegative term?

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

**Answer
5** :

(i) Given:

AP: 24, 23 ¼, 22 ½, 21¾, … = 24, 93/4, 45/2, 87/4, …

Here, a_{1} =a = 24, a_{2} = 93/4

Common difference, d =a_{2} – a_{1} = 93/4 – 24

= (93 – 96)/4

= – 3/4

We know, a_{n} =a + (n – 1) d [where a is first term or a_{1} and d is commondifference and n is any natural number]

We know, a_{n} =a + (n – 1) d

a_{n} =24 + (n – 1) (-3/4)

= 24 – 3/4n + ¾

= (96+3)/4 – 3/4n

= 99/4 – 3/4n

Now we need to find,first negative term.

Put a_{n} <0

a_{n} =99/4 – 3/4n < 0

99/4 < 3/4n

3n > 99

n > 99/3

n > 33

Hence, 34^{th} termis the first negative term of given AP.

(ii) Given:

AP: 12 + 8i, 11 + 6i,10 + 4i, …

Here, a_{1} =a = 12 + 8i, a_{2} = 11 + 6i

Common difference, d =a_{2} – a_{1}

= 11 + 6i – (12 + 8i)

= 11 – 12 + 6i – 8i

= -1 – 2i

We know, a_{n} =a + (n – 1) d [where a is first term or a_{1} and d is commondifference and n is any natural number]

a_{n} =12 + 8i + (n – 1) -1 – 2i

= 12 + 8i – n – 2ni +1 + 2i

= 13 + 10i – n – 2ni

= (13 – n) + (10 – 2n)i

To find purely realterm of this A.P., imaginary part have to be zero

10 – 2n = 0

2n = 10

n = 10/2

= 5

Hence, 5^{th} termis purely real.

To find purelyimaginary term of this A.P., real part have to be zero

∴ 13 – n = 0

n = 13

Hence, 13^{th} termis purely imaginary.

(i) How many terms are in A.P. 7, 10, 13,…43?

(ii) How many terms are there in the A.P. -1, -5/6, -2/3, -1/2, …, 10/3 ?

**Answer
6** :

(i) Given:

AP: 7, 10, 13,…

Here, a_{1} =a = 7, a_{2} = 10

Common difference, d =a_{2} – a_{1} = 10 – 7 = 3

We know, a_{n} =a + (n – 1) d [where a is first term or a_{1} and d is commondifference and n is any natural number]

a_{n} = 7+ (n – 1)3

= 7 + 3n – 3

= 3n + 4

To find total terms ofthe A.P., put a_{n} = 43 as 43 is last term of A.P.

3n + 4 = 43

3n = 43 – 4

3n = 39

n = 39/3

= 13

Hence, total 13 termsexists in the given A.P.

(ii) Given:

AP: -1, -5/6, -2/3,-1/2, …

Here, a_{1} =a = -1, a_{2} = -5/6

Common difference, d =a_{2} – a_{1}

= -5/6 – (-1)

= -5/6 + 1

= (-5+6)/6

= 1/6

_{n} =a + (n – 1) d [where a is first term or a_{1} and d is commondifference and n is any natural number]

a_{n} =-1 + (n – 1) 1/6

= -1 + 1/6n – 1/6

= (-6-1)/6 + 1/6n

= -7/6 + 1/6n

To find total terms ofthe AP,

Put a_{n} =10/3 [Since, 10/3 is the last term of AP]

a_{n} =-7/6 + 1/6n = 10/3

1/6n = 10/3 + 7/6

1/6n = (20+7)/6

1/6n = 27/6

n = 27

Hence, total 27 termsexists in the given A.P.

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

**Answer
7** :

Given:

First term, a = 5;last term, l = a_{n} = 80

Common difference, d =3

_{n} =a + (n – 1) d [where a is first term or a_{1} and d is commondifference and n is any natural number]

a_{n} = 5+ (n – 1)3

= 5 + 3n – 3

= 3n + 2

To find total terms ofthe A.P., put a_{n} = 80 as 80 is last term of A.P.

3n + 2 = 80

3n = 80 – 2

3n = 78

n = 78/3

= 26

Hence, total 26 termsexists in the given A.P.

The 6^{th} and 17^{th} terms of an A.P. are 19and 41 respectively. Find the 40^{th} term.

**Answer
8** :

Given:

6^{th} termof an A.P is 19 and 17^{th} terms of an A.P. is 41

So, a_{6} =19 and a_{17} = 41

_{n} =a + (n – 1) d [where a is first term or a_{1} and d is commondifference and n is any natural number]

When n = 6:

a_{6} = a+ (6 – 1) d

= a + 5d

Similarly, When n= 17:

a_{17} =a + (17 – 1)d

= a + 16d

According to question:

a_{6} =19 and a_{17} = 41

a + 5d = 19 ……………… (i)

And a + 16d = 41…………..(ii)

Let us subtractequation (i) from (ii) we get,

a + 16d – (a + 5d) =41 – 19

a + 16d – a – 5d = 22

11d = 22

d = 22/11

= 2

put the value of d inequation (i):

a + 5(2) = 19

a + 10 = 19

a = 19 – 10

= 9

As, a_{n} =a + (n – 1)d

a_{40} =a + (40 – 1)d

= a + 39d

Now put the value of a= 9 and d = 2 in a_{40} we get,

a_{40} =9 + 39(2)

= 9 + 78

= 87

Hence, 40^{th} termof the given A.P. is 87.

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

**Answer
9** :

Given:

9^{th} termof an A.P is 0

So, a_{9} =0

We need to prove: a_{29} =2a_{19}

_{n} =a + (n – 1) d [where a is first term or a_{1} and d is commondifference and n is any natural number]

When n = 9:

a_{9} = a+ (9 – 1)d

= a + 8d

According to question:

a_{9} = 0

a + 8d = 0

a = -8d

When n = 19:

a_{19} =a + (19 – 1)d

= a + 18d

= -8d + 18d

= 10d

When n = 29:

a_{29} =a + (29 – 1)d

= a + 28d

= -8d + 28d[Since, a = -8d]

= 20d

= 2×10d

a_{29} =2a_{19} [Since, a_{19} = 10d]

Hence Proved.

If 10 times the 10^{th} term of an A.P. is equal to 15 timesthe 15^{th} term, show that the 25^{th} term of theA.P. is Zero.

**Answer
10** :

Given:

10 times the 10^{th} termof an A.P. is equal to 15 times the 15^{th} term

So, 10a_{10} =15a_{15}

We need to prove: a_{25} =0

_{n} =a + (n – 1) d [where a is first term or a_{1} and d is commondifference and n is any natural number]

When n = 10:

a_{10} =a + (10 – 1)d

= a + 9d

When n = 15:

a_{15} =a + (15 – 1)d

= a + 14d

When n = 25:

a_{25} =a + (25 – 1)d

= a + 24d ………(i)

According to question:

10a_{10} =15a_{15}

10(a + 9d) = 15(a +14d)

10a + 90d = 15a + 210d

10a – 15a + 90d – 210d= 0

-5a – 120d = 0

-5(a + 24d) = 0

a + 24d = 0

a_{25} =0 [From (i)]

Hence Proved.

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