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If a,b,c are in A.P. Prove that b+c,c+a,a+b are also in A.P.?

If a,b,c are in A.P. Prove that b+c,c+a,a+b are also in A.P.?

Grade:12th pass

11 Answers

AKASH GOYAL AskiitiansExpert-IITD
420 Points
13 years ago
Dear Rohan
If a,b,c are in AP then 2b=a+c
We have to prove
2(a+c)=(b+c)+(a+b)
so 2a+2c=2b+a+c
so a+c=2b
Which is true
Hence proved

All the best.
AKASH GOYAL
AskiitiansExpert-IITD

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vikas askiitian expert
509 Points
13 years ago

 if a,b,c are in AP then

  b-a=c-b

   a+c=2b ...........1

now we have to prove that b+c,c+a,a+b are in AP,

     let these are in AP then

                 (a+c)-(b+c)=(a+b)-(a+c)

                      (a+c)=2b

from eq 1 we have a+c=2b ,RHS=LHS

so these terms are also in AP...

 

Poornima
19 Points
7 years ago
a,b,c. Are in apSo their common difference will be equalSo,b-a=c-ba+c=2bTo proveb+c,c+a,a+b are in apProofLet us consider that they are in apSo,(C+a)-(b+c)=(a+b)-(c+a)c+a-b-c=a+b-c-aa-b=b-ca+c=2bTheir common difference is equal.So they are in ap
Ashutosh Kumar
19 Points
7 years ago
Solution-a,b&c are in ap .Now , a+(-a-b-c),b+(-a-b-c)& c+(-a-b-c) are in ap.Now,-b-c,-a-c&-a-b are in apOr,-(b+c),-(a+c)&-(a+b)are in apor,b+c,a+c&a+b are in ap
Ashutosh Bajpai
13 Points
7 years ago
If b+c ,c+a,a+b are in AP then c+a -(b+c)=a+b-(c+a)a-b =b-cThen 2b=a+cThus it proved that a,b and c are in Ap
vallish
11 Points
7 years ago
here a b c are in ap then
b-a=c-b take – common
-(a-b)=-(b-c) divide both by -
a-b=b-c.........(1)
assume a+b b+c a+c are ap
b+c -(a+b)=a+c-(b+c)
a-b=b-c.........(2)
from 1 & 2 hence proof
Ritika Das
105 Points
7 years ago
Well, since it is given that a,b and c are in Arithmetic Progression, for any arbitrary common difference d:-
b=a+d ;
c=a+2d.
Therefore, 
a+b=2a+d;
a+c=2a+2d;
b+c=2a+3d.
Hete all the expressions have a common difference d.
Thus proved that a+b, a+c and b+c are in Arithmetic Progression for the common difference d.
Prabhleen kaur
24 Points
7 years ago
If a,b,c are in AP. So,b-a=c-b0=c+a-b-b0=c+a-2b2b=c+a ..................1Let us assume that b+c, c+a, a+b are in AP.So, (c+a)-(b+c) = (a+b)-(c+a)(a+c)=2b2b=2b [from eq 1]LHS=RHSTherefore, b+c, c+a and a+b are in AP. Hence,proved
Prabhleen kaur
24 Points
7 years ago
A+c =2b.......1Let us assume that b+c,c+a and a+b are in AP(c+a)-(b+c)=(a+b)-(c+a)a+c=2b2b=2b[from eq 1]LHS=RHSTHEREFORE B+C,C+A AND A+B ARE IN AP.hence, proved
ARKADEEP DAS
44 Points
7 years ago
let common differene be d
then b=a+d
c=a+2d
now a+c = 2a+2d
b+c=2a+3d
a+b=2a+d
so see there are in ap
ok rohan u are done
Kushagra Madhukar
askIITians Faculty 628 Points
4 years ago
Dear student,
Please find the solution to your problem below.
 
If a,b,c are in AP then
b – a = c – b
a + c = 2b
Adding, a + c on both sides
Hence, 2(a + c) = 2b + a + c
or, 2(a + c) = b + a + b + c
or, 2(a + c) = (a + b) + (b + c)
Hence, (b + c), (c + a), (a + b) are also in AP
 
Hope it helps.
Thanks and regards,
Kushagra

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