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

rohan kanojia, 10 years ago

Grade:12th pass

11 Answers

AKASH GOYAL AskiitiansExpert-IITD

419 Points

10 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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Approved

vikas askiitian expert

509 Points

10 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...

Approved

Poornima

19 Points

4 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

4 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

4 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

4 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

3 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

3 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

3 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