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### Math Assignment Class XI Ch -09 | Sequence & Series

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## Math Assignment Class XI Ch -09

**SEQUENCE & SERIES**

**Extra questions of chapter 09 sequence & series class 11 with answers and hints to the difficult questions, strictly according to the CBSE and DAV Board. Important and useful math. assignment for the students of class 11**

**MATHEMATICS ASSIGNMENT ON **

**SEQUENCE AND SERIES (XI)**

**Strictly according to the CBSE and DAV Board**

Question 1

If 9 times the 9th term of an AP is equal to 13 times the 13th term, then find 22nd term of the AP?

Ans: Zero

Question 2

If
a and b are the roots of x^{2} - 3x + p = 0 and c, d are the roots of x^{2
}- 12x + q = 0, where a, b, c, d
forma G.P. Prove
that (q + p) : (q - p) = 17: 15.

Solution Hint: r^{2} = 4

Question 3

Between
5 and 35, m numbers have been inserted in such a way that the resulting
sequence is an A.P. and the ratio of 3rd and (m - 2)th numbers is 7 : 13. Find
the value of m.

Ans:
m = 9

Question 4

If
the first term of GP is a and nth term is b and P denotes the product of first
n terms. Prove that P^{2} = (ab)^{2}

Question 5

If
sum of n terms of two AP’s are in the ratio
3n + 8 : 7n + 15, then find the ratio of their 12^{th} terms

Ans: 77/176

Question 6

Find the sum of n terms of the series

3 + 5 + 9 + 15 + 23 + …………………. N terms

Ans: n^{2}+2n

Question 7

The product of the first three terms of a G.P.
is 1000. If we add 6 to its second term and 7 to its third term, the resulting
three terms form an A.P. Find the terms of the G.P.

Ans: 5, 10, 20 or 20, 10, 5

Question 8

If a, b and c are in A.P and ab + bc + ca = 0
then prove that a²(b + c), b²(a + c), c^{2}(a + b) are also in A.P.

Question 9

Find the sum of the following series up to n terms :

Ans:

Question 10

The ratio of
sums of m and n terms of an A.P. is m^{2} : n^{2}. Show that
ratio of m^{th} and n^{th} term is (2m – 1) : (2n – 1)

Question 11

The sum of two positive numbers is 4 times their geometric mean, show that numbers are in the ratio (2 + √3 ) : (2 – √3 ).

OR

Find the ratio of two positive numbers a and b such that AM: GM = 2 : 1

Ans: (2 + √3 ) : (2 – √3 ).

OR

The AM of two positive numbers is 3 times their GM. Prove that numbers are in the ratio (3+2√2) : (3-2√2)

Question12

Find the sum of n terms of the series:

0.6 + 0.66 + 0.666 + ..............

Ans:

Question 13

Aman saw his younger brother playing with building blocks. he observed that he is playing in a pattern by making a tower of 32 blocks and then dividing into half a tower of 16 blocks and so on still he got a tower of 1 block.

a) What kind of pattern is brother making? Also, how many Towers did the boy make?

Ans: Brother making the towers of blocks: 32, 16, 8, 4, 2, 1 and it is in GP

b) How many total blocks does the boy carry with him?

Ans: 63

Solution Hint

Total towers made by the boy = 6

Here a = 32, r = 1/2, n = 6, So find S_{6}

Question 14

On a Sunday Sunil and his friend went to see a circus show and observed
that the arrangement of chairs in a row has a Peculiar arrangement. The number
of chairs in his row, which is the third row, are 125 and the number of chairs
in the First row are 75. The number of chairs is increasing by a fixed number.
There are 12 rows of chairs in all.

a) How many chairs are there in the 7th row, where his friend is situated?

Ans: a = 75, a_{3} = 125
d = 25

a_{7} = a + 6d = 225

b) If all the chairs are occupied, how many people saw that show? also
find the Arithmetic Mean of the number of chairs in the row Sunil and his
friend was seated?

Ans: S12 = 2550

AM of 125 and 225 = 175

**Question 17**

In a cricket tournament 16 school teams participated. A sum of ` 8000
is to be awarded among themselves as prize money. If the last placed
team is awarded ` 275 in prize money and the award increases by the
same amount for successive finishing places, how much amount will the
first place team receive?

Ans: ₹ 725

**Question 18**

Find the r^{th}
term of an AP sum of whose first n terms is 2n + 3n^{2}.

Ans: 6r - 1

**Question 19**

If the third term of GP is 4, then the product of its first 5 terms is

Ans: (4)^{5}.

Solution Hint: It is given that, T_{3} = 4

Let a and r the first term and common ratio, respectively.

Then, ar^{2} =
4 ...(i)

Product of first 5 terms = ⋅ ⋅ ⋅ ⋅ a (ar) (ar^{2} ) (ar^{3} ) (ar^{4} )

= a^{5} r^{10 }= (ar^{2})^{5} = (4)^{5}. ....... using (i)

**Question 20**

If S_{n} denotes the sum of first n terms of an AP, If S_{2n} = 3S_{n}, then find S_{3n} : S_{n}

Ans: 6 : 1

**Question 21**

In a set of four numbers, the first three are in GP and the last three are in AP with a common difference of 6. If the first and fourth terms are equal, find the four numbers.

Ans: 8, - 4, 2, 8

Explanation: As the last three numbers are in A.P. with a common difference of 6

Let the last three numbers be b, b + 6 and b + 12.

Let a be the first number, then the four numbers in order are a, b, b + 6 and b + 12.

According to the question, the first three numbers are in G.P.

b^{2} = a (b + 6) …….. (1)

Also first and fourth numbers are equal, so

a = b + 12 …………. (ii)

Substituting the value of a from (ii) in (i), we get

b^{2} = (b + 12) (b + 6)

b^{2} = b^{2 }+ 18b + 72

18b + 72 = 0

b = - 4

From (ii), we get a = - 4 + 12 = 8

Hence, the required numbers are 8, - 4, 2, 8** **

**Question 22**

If the sum of an infinite geometric series is 15 and the sum
of the squares of the terms of the series is 45, find the G.P.

Ans:

Solution Hint:

Using the formula for sum to infinite terms

ATQ

sum of an infinite geometric series is 15

⇒ a = 15 - 15r .......(i)

sum of the squares of the infinite terms of the series is 45,

**THANKS FOR YOUR VISIT**

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