Test: Introduction to Arithmetic Progression - Class 10 MCQ

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.

as there are 11 students  so 6 th student would be middle one
a=45
a+(6-1)d=55
a+5d=55
45+5d=55
5d=10
d=2

We have the sequence 5n²+2n+3
It will be in AP if it satisfies an-an-1=d where d is a constant.
Here, a_n=5n²+2n+3
And, a_n-1=5(n-1)²+2(n-1)+3=5(n²+1-2n)+2n-2+3
=5n²+5-10n+2n+1=5n²-8n+6
a_n-a_n-1=5n²+2n+3-(5n²-8n+6)
=5n²+2n+3-5n²+8n-6
=10n-3, which depends on a variable , so its not constant.

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

A¹=5 d=2
a²=a¹+d =5+2=7
a³=a²+d =7+2 =9
a⁴=a³+d=9+2=11
A⁵ =a⁴+d=11+2=13
Therefore 5,7,9,11,13 is an AP.

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

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

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

On 1st Monday he did 25
on 2nd next he did 25+5=30 
on 3rd Monday he did 30+5=35

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

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

AP is 
So a = 
So 

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