Question

Any 10th grader solve it
for 50 points​

Answer 1

`(a)/(p)* (q-r)+(b)/(q)* (r-p)+(c)/(r)* (p-q)\\eq 0` is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie.,
`a_(1)=A` and common difference=D

The nth term can be written as

`a_(n)=A+(n-1)D`

pth term of given arithmetic progression is a

`a_(p)=A+(p-1)D=a`

qth term of given arithmetic progression is b

`a_(q)=A+(q-1)D=b` and

rth term of given arithmetic progression is c

`a_(r)=A+(r-1)D=c`

We have to prove that

`(a)/(p)* (q-r)+(b)/(q)* (r-p)+(c)/(r)* (p-q)=0`

Now to prove LHS=RHS

Now take LHS

`(a)/(p)* (q-r)+(b)/(q)* (r-p)+(c)/(r)* (p-q)`

`=(A+(p-1)D)/(p)* (q-r)+(A+(q-1)D)/(q)* (r-p)+(A+(r-1)D)/(r)* (p-q)`

`=(A+pD-D)/(p)* (q-r)+(A+qD-D)/(q)* (r-p)+(A+rD-D)/(r)* (p-q)`

`=(Aq+pqD-Dq-Ar-prD+rD)/(p)+(Ar+rqD-Dr-Ap-pqD+pD)/(q)+(Ap+prD-Dp-Aq-qrD+qD)/(r)`

`=([Aq+pqD-Dq-Ar-prD+rD]* qr+[Ar+rqD-Dr-Ap-pqD+pD]* pr+[Ap+prD-Dp-Aq-qrD+qD]* pq)/(pqr)`

`=(Arq^(2)-Dq^(2)r-Aqr^(2)+qr^(2)D+Apr^(2)-pDr^(2)-Ap^(2)r+p^(2)rD+Ap^(2)q-Dp^(2)q-Aq^(2)p+q^(2)pD)/(pqr)`

`=(Arq^(2)-Dq^(2)r-Aqr^(2)+qr^(2)D+Apr^(2) -pDr^(2)-Ap^(2)r+p^(2)rD+Ap^(2)q-Dp^(2)q-Aq^(2)p+q^(2)pD)/(pqr)`

`\\eq 0`

ie.,
`RHS\\eq 0`

Therefore
`LHS\\eq RHS`

ie.,
`(a)/(p)* (q-r)+(b)/(q)* (r-p)+(c)/(r)* (p-q)\\eq 0`

Hence proved