Chapter 9 Sequences and Series Ex-9.1 |
Chapter 9 Sequences and Series Ex-9.3 |
Chapter 9 Sequences and Series Ex-9.4 |

**Answer
1** :

The odd integers from1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms asequence in A.P.

Where, the firstterm, *a* = 1

Commondifference, *d* = 2

Now,

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n – 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

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

Therefore, the sum ofodd numbers from 1 to 2001 is 1002001.

**Answer
2** :

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

It clearly forms a sequence in A.P.

Where, the first term, a = 105

Common difference, d = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n – 5 = 995

5n = 995 – 105 + 5 = 895

n = 895/5

n = 179

We know,

Sn = n/2 [2a + (n-1)d]

Therefore, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

**Answer
3** :

Given,

The first term (a) ofan A.P = 2

Let’s assume *d* bethe common difference of the A.P.

So, the A.P. will be2, 2 + *d*, 2 + 2*d*, 2 + 3*d*, …

Then,

Sum of first fiveterms = 10 + 10*d*

Sum of next five terms= 10 + 35*d*

From the question, wehave

10 + 10d = ¼ (10 +35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a_{20} =a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112

Therefore, the 20^{th} termof the A.P. is –112.

**Answer
4** :

Let’s consider the sum of n terms of the given A.P. as –25.

We known that,

Sn = n/2 [2a + (n-1)d]

where n = number of terms, a = first term, and d = common difference

So here, a = –6

d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2

Thus, we have

**Answer
5** :

**Answer
6** :

Given A.P.,

25, 22, 19, …

Here,

First term, a = 25 and

Common difference, d =22 – 25 = -3

Also given, sum ofcertain number of terms of the A.P. is 116

The number of terms ben

So, we have

S_{n} =n/2 [2a + (n-1)d] = 116

116 = n/2 [2(25) +(n-1)(-3)]

116 x 2 = n [50 – 3n +3]

232 = n [53 – 3n]

232 = 53n – 3n^{2}

3n^{2} –53n + 232 = 0

3n^{2} –24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8)= 0

(3n – 29) (n – 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be anintegral value, n = 8

Thus, 8^{th} termis the last term of the A.P.

a_{8} =25 + (8 – 1)(-3)

= 25 – 21

= 4

**Answer
7** :

Given, the *k*^{th} termof the A.P. is 5*k* + 1.

*k*^{th} term = *a _{k}* =

And,

*a *+ (*k* – 1)*d* = 5*k* +1

*a* + *kd* – *d* =5*k* + 1

On comparing thecoefficient of *k*, we get *d* = 5

*a *– *d *= 1

*a* – 5 = 1

⇒ *a* =6

If the sum of *n* terms of an A.P. is (*pn *+ *qn*^{2}),where *p* and *q* are constants, find the commondifference.

**Answer
8** :

We know that,

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

From the question wehave,

On comparing thecoefficients of *n*^{2} on both sides, we get

d/2 = q

Hence, *d* =2*q*

Therefore, the commondifference of the A.P. is 2*q*.

**Answer
9** :

Let *a*_{1}, *a*_{2},and *d*_{1}, *d*_{2 }be the firstterms and the common difference of the first and second arithmetic progressionrespectively.

Then, from thequestion we have

**Answer
10** :

Let’s take *a* and *d* tobe the first term and the common difference of the A.P. respectively.

Then, it given that

Therefore, the sum of(p + q) terms of the A.P. is 0.

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