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Chapter 9 Sequences and Series Ex-9.2 Interview Questions Answers

Question 1 : Find the sum of odd integers from 1 to 2001.

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,

Sn =n/2 [2a + (n-1)d]

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

Question 2 : Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

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.

Question 3 : In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.

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 + 2d, 2 + 3d, …

Then,

Sum of first fiveterms = 10 + 10d

Sum of next five terms= 10 + 35d

From the question, wehave

10 + 10d = ¼ (10 +35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a20 =a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112

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

Question 4 : How many terms of the A.P. -6, -11/2, -5, …. are needed to give the sum –25?

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

Question 5 : In an A.P., if pth term is 1/q and qth term is 1/p, prove that the sum of first pq terms is ½ (pq + 1) where p ≠ q.

Answer 5 :


Question 6 : If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term

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

Sn =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 – 3n2

3n2 –53n + 232 = 0

3n2 –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, 8th termis the last term of the A.P.

a8 =25 + (8 – 1)(-3)

= 25 – 21

= 4

Question 7 : Find the sum to n terms of the A.P., whose kth term is 5k + 1.

Answer 7 :

Given, the kth termof the A.P. is 5k + 1.

kth term = ak = +(k – 1)d

And,

+ (k – 1)d = 5k +1

a + kd – d =5k + 1

On comparing thecoefficient of k, we get d = 5

– = 1

a – 5 = 1

 a =6

Question 8 :

If the sum of n terms of an A.P. is (pn qn2),where p and q are constants, find the commondifference.

Answer 8 :

We know that,

Sn =n/2 [2a + (n-1)d]

From the question wehave,

On comparing thecoefficients of n2 on both sides, we get

d/2 = q

Hence, d =2q

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

Question 9 : The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms.

Answer 9 :

Let a1a2,and d1dbe the firstterms and the common difference of the first and second arithmetic progressionrespectively.

Then, from thequestion we have

Question 10 : If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.

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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Chapter 9 Sequences and Series Ex-9.2 Contributors

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