### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

## Math Assignment  Class XI  Ch -09SEQUENCE & 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

## 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 x2 - 3x + p = 0 and c, d are the roots of x2 - 12x + q  = 0, where a, b, c, d forma G.P. Prove that (q + p) : (q - p) = 17: 15.

Solution Hint:  r2 = 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 P2 = (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 12th terms

Ans:  77/176

Question 6

Find the sum of n terms of the series

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

Ans: n2+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), c2(a + b) are also in A.P.

Question 9

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

$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}\: + ......$

Ans:  $\frac{n}{25}\left ( 2n^{2}+9n+13 \right )$

Question 10

The ratio of sums of m and n terms of an A.P. is m2 : n2. Show that ratio of mth and nth 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:  $\frac{2n}{3}-\frac{2}{27}(1-10^{-n})$

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

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, a3 = 125    d = 25

a7 = 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 15
The sum of three numbers in GP is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an AP . Then find the numbers.
Ans: 8, 16, 32, or 32, 16, 8

Question 16
Find the sum of first 20 terms of the series

$1+\frac{1+2}{2}+\frac{1+2+3}{3}+\frac{1+2+3+4}{4}+ ........$

Ans: 115

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 rth term of an AP sum of whose first n terms is 2n + 3n2.

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, T3 = 4

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

Then, ar2 = 4 ...(i)

Product of first 5 terms =     a (ar) (ar2 ) (ar3 ) (ar4 )

= a5 r10   = (ar2)5 = (4)5.  ....... using (i)

Question 20

If Sn denotes the sum of first n terms of an AP, If S2n = 3Sn, then find S3n : Sn

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.

b2 = 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

b2 = (b + 12) (b + 6)

b2 = b+ 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:    $5,\frac{10}{3},\frac{20}{9},\frac{40}{27}$

Solution Hint:

Using the formula for sum to infinite terms

$S_{\infty }=\frac{a}{1-r}$

ATQ

sum of an infinite geometric series is 15

$\frac{a}{1-r}=15$

⇒ a = 15 - 15r .......(i)

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

$\frac{a^{2}}{1-r^{2}}=45$

$\Rightarrow \left ( \frac{a}{1-r} \right )\left ( \frac{a}{1+r} \right )=45$

$\Rightarrow 15\left ( \frac{a}{1+r} \right )=45$

$\Rightarrow \left ( \frac{a}{1+r} \right )=3$

⇒ a = 3 + 3r  ........ (ii)

Solving eqn. (i) and eqn. (ii) we get

r = 2/3  and  a = 5
So required GP is   $5,\frac{10}{3},\frac{20}{9},\frac{40}{27}$

THANKS FOR YOUR VISIT