Question

If a,b,c are in arithmetic progression , then show that the following are in arithematic progression :-

1. b+c , c+a , a+b

2. a+3k , b+3k , c+3k

3. a\7k , b\7k , c\7k

4. b+c-a , c+a-b , a+b-c

Answer 1

Let a be the first term
And d be thecommon difference
Then a=a
b=a+d
C=a+2d. (Since a,b,c are in arithmetic p
Progression

1) b+c=(a+d )+(a+2d)=2a+3d
C+a=(a+2d)+a=2a+2d
a+b=a+(a+d)=2a+d
Here
(b+c)-(c+a)=-d
(a+b)-(c+a)=-d
Thus
b+c ,c+a,a+b are in arithmetic
With -d ascommon difference
And 2a+3d first number


2)a+3k=a+3k
b+3k=(a+d)+3k=a+d+3k
C+3k=(a+2d)+3k=a+2d+3k
(b+3k)-(a+3k)=d
(c+3k)-(b+3k)=d
So these 3 are in arithmetic with d as common differnce and a+3k as first number

​​​​​3)a/7k
b/7k=(a+d)/7k
​​​​​ c/7k=(a+2d)/7k
(b/7k/)-(a/7k)=​d/k​​​​​​
(C/7k)-(b/7k)=d/k
So these 3 are in arithmetic withd
as d/7 as common difference and a/7k as starting number

4)b+c-a=(a+d)+(a+2d)-a
=a+3d
C+a-b=(a+2d)+a-(a+d)
=a+d
a+b-c=a+(a+d)-(a+2d)
=a-d
These 3 are in arithemtic with -2d as common difference and
a first number