Answer:
[tex]\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0[/tex] is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.
Step-by-step 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., [tex]a_{1}=A[/tex] and common difference=D
The nth term can be written as
[tex]a_{n}=A+(n-1)D[/tex]
pth term of given arithmetic progression is a
[tex]a_{p}=A+(p-1)D=a[/tex]
qth term of given arithmetic progression is b
[tex]a_{q}=A+(q-1)D=b[/tex] and
rth term of given arithmetic progression is c
[tex]a_{r}=A+(r-1)D=c[/tex]
We have to prove that
[tex]\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0[/tex]
Now to prove LHS=RHS
Now take LHS
[tex]\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)[/tex]
[tex]=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)[/tex]
[tex]=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)[/tex]
[tex]=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}[/tex]
[tex]=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}[/tex]
[tex]=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}[/tex]
[tex]=\frac{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}[/tex]
[tex]=\frac{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}[/tex]
[tex]\neq 0[/tex]
ie., [tex]RHS\neq 0[/tex]
Therefore [tex]LHS\neq RHS[/tex]
ie.,[tex]\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0[/tex]
Hence proved
