Question

if x, y, z be respectively the pth, qth and rth term of a G.P., Show that (q-r) log x+(r-p) log y+(p-q) log z=0​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{x,y,z\,\,\,are\,\,\,respectively\,\,\,the\,\,\,p^{th},\,q^{th},\,r^{th}\,\,\,term\,\,\,of\,\,\,a\,\,GP}$

Let the first term of the GP be 'A' and the common ratio be 'R'

So,

$\sf{x=A\,R^{p-1},\,\,y=A\,R^{q-1},\,\,z=A\,R^{r-1}}$

$\sf{\implies\,log(x)=log(A\,R^{p-1}),\,\,log(y)=log(A\,R^{q-1}),\,\,log(z)=log(A\,R^{r-1})}$

$\sf{\implies\,log(x)=log(A)+log(R^{p-1}),\,\,log(y)=log(A)+log(R^{q-1}),\,\,log(z)=log(A)+log(R^{r-1})}$$\sf{\implies\,log(x)=log(A)+(p-1)log(R)}\\\sf{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,log(y)=log(A)+(q-1)log(R)}\\\sf{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,log(z)=log(A)+(r-1)log(R)}$

Now,

$\sf{\implies\,(q-r)log(x)=(q-r)log(A)+(q-r)(p-1)log(R)}\\\sf{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(r-p)log(y)=(r-p)log(A)+(r-p)(q-1)log(R)}\\\sf{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(p-q)log(z)=(p-q)log(A)+(p-q)(r-1)log(R)}$

$\sf{\implies\,(q-r)log(x)=(q-r)log(A)+(pq-pr-q+r)log(R)}\\\sf{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(r-p)log(y)=(r-p)log(A)+(qr-pq-r+p)log(R)}\\\sf{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(p-q)log(z)=(p-q)log(A)+(pr-qr-p+r)log(R)}$

Add the above expressions

And we get,

$\sf{\implies\,(q-r)log(x)+(r-p)log(y)+(p-q)log(z)=0}$