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QUESTION: 1

Roots of quadratic equation *x*^{2} – 3*x* = 0, will be

Solution:

QUESTION: 2

For an A.P. the third and the fifth terms are given as 10 and 16 .What is the fourth term and the common difference?

Solution:

QUESTION: 3

If p, q, r, s, t are the terms of an A.P. with common difference -1 the relation between p and t is:

Solution:

p, q, r, s, t are the terms of AP. So a=p , d= -1

To find a relation between t and p we use

l=a+(n-1)d, where l=t and n=5

t=p+(5-1)(-1)=p+4(-1)=p-4

This is because the difference is negative so it will be smaller than p.

QUESTION: 4

The weights of 11 students selected for a team are noted in ascending order and are in A. P. The lowest value is 45 Kg, and the middle value is 55 Kg. What is the difference between the two values placed consecutively ?

Solution:

QUESTION: 5

Is the sequence, whose general term is 5n2 + 2n + 3 an AP?

Solution:

We have the sequence 5n^{2}+2n+3

It will be in AP if it satisfies an-an-1=d where d is a constant.

Here, a_{n}=5n^{2}+2n+3

And, a_{n-1}=5(n-1)^{2}+2(n-1)+3=5(n^{2}+1-2n)+2n-2+3

=5n^{2}+5-10n+2n+1=5n^{2}-8n+6

a_{n}-a_{n-1}=5n^{2}+2n+3-(5n^{2}-8n+6)

=5n^{2}+2n+3-5n^{2}+8n-6

=10n-3, which depends on a variable , so its not constant.

QUESTION: 6

Find the next two terms of the A.P.:- -10, -6,-2…

Solution:

We have A.P.: -10, -6,-2…

So a= -10, d=-6+10=-2+6=4

l=a+(n-1)d

For fourth term,

= -10+(3)*4=-10+12=2

For 5th term

= -10+4*4=6

QUESTION: 7

If a + 1, 2a + 1, 4a – 1 are in A.P., then value of ‘a’ is:

Solution:

QUESTION: 8

Give that an A.P. has term as 5 and common differences as 2. What is the A.p.?

Solution:

A¹=5 d=2

a²=a¹+d =5+2=7

a³=a²+d =7+2 =9

a⁴=a³+d=9+2=11

A^{5} =a⁴+d=11+2=13

Therefore 5,7,9,11,13 is an AP.

QUESTION: 9

The 5th term of an A.P. is 18, the common difference is 2. What is the first term ?

Solution:

5th term of AP is 18 and common difference is 2

We have, l = a + (n - 1)d

Where l = 18,a = ?,n = 5,d = 2

Substituting the values,

18 = a+8

a = 10

QUESTION: 10

Ramesh’s salary in February 2008 is Rs. 10,000. If he’s promised an increase of Rs. 1000 every year, what would be his salary in Feb 2011?

Solution:

So this is an AP with a=10,000 , d= 1000

So after 4 years

l=a+(n-1)d=10000+(4 -1)1000=10000+3000=13000

QUESTION: 11

Given an A.P. few of whose terms are x, y, 2, 4, 6, 8,………. What must be the values of x and y?

Solution:

Arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant,d

AP is x, y, 2, 4, 6, 8,…

So we have 4-2=2,6-4=2 so we have d=2

So y=2-2=0

And x=0-2=-2

QUESTION: 12

If p – 1, p + 3, 3p – 1 are in A.P., then p is equal to:

Solution:

We have given that

p-1,p+3 , 3p-1 are in A.P

we have to find p= ?

solution :-

we know that :

if a,b,c are in AP

then 2b = a + c

here

=> 2(p+3) = { (p-1) + ( 3p -1) }

=> 2p +6 = 4p -2

=> 2p = 8

=> p = 4

QUESTION: 13

Which of the following is an A.P with -10 as first term and 2 as the common difference?

Solution:

QUESTION: 14

Amit starts his exercise regime with 25 push ups on Monday. He plans to increase 5 push ups every following Monday. How many push ups will he be doing on the 3rd Monday since he started?

Solution:

On 1st Monday he did 25

on 2nd next he did 25+5=30

on 3rd Monday he did 30+5=35

QUESTION: 15

For what values of k can the following numbers form an A.P:2k-1,k+2,2k

Solution:

If 2k-1,k+2,2k are in AP, the difference of two consecutive terms will be same.

Therefore,

QUESTION: 16

From an A.P. 5 consecutive terms are required to be written. The central term out of the five terms is given as 5 and it’s informed that the common difference of the A.P. is 4. Write the 5 terms in the correct sequence.

Solution:

AP is: a,a+d,a+2d,a+3d,a+4d

Middle term is a+2d=5

d=4

a+2d=5

a+2*4=5

a=5-8= -3

a+d= -3+4=1

a+3d= -3+3*4=-3+12=9

a+4d= -3+4*4= -3+16=13

So the terms are : -3,1,5,9,13

QUESTION: 17

If 4/5, a, 2 are three consecutive terms of an A.P., then the value of a is:

Solution:

QUESTION: 18

If a - b, 0 and a + b are consecutive terms of an AP then

Solution:

QUESTION: 19

The next term of the A.P. … is:

Solution:

AP is

So a =

So

QUESTION: 20

The angles of a triangle in A.P. the smallest being half of the greatest. So what are the angles ?

Solution:

Arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant,d.

So let the angles be a-d , a and a+d

We are given smallest angle is half of the largest

So, a-d=1/2(a+d)

2(a-d)=a+d

2a-2d=a+d

a=3d

Using angle sum property,

(a-d)+a+(a+d)=180

3a=180

a=60

a=3d

3d=60

d=20

a-d=60-20=40

a=60

a+d=60+20=80

So , the angles are 40,60,80

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