If `p^(t h),\ q^(t h),\ r^(t h)a n d\ s^(t h)` terms of an A.P. are in G.P., then show that `(p-q),\ (q-r),\ (r-s)` are also in G.P.

Explanation

We know that the `n^(th)` term of AP is `a+(n-1)d`

i.e. `a_n=a+(n-1)d`


`(##XI_09_SLV_21_S01##)`

It is given that `a_p,a_q,a_r,a_s` in GP

i.e. their common ratio is same

So, `a_q/a_p=a_r/a_q=a_s/a_r` ` " " " " .....(1)`

We need to show that`(p - q)`, `(q - r)`, `(r - s)`are also in G.P.

i.e. we need to show their common ratio is same

i.e. we need to show `(q-r)/(p-q)=(r-s)/(q-r)`

It is given `a_q/a_p=a_r/a_q=a_s/a_r`


Taking `a_q/a_p=a_r/a_q`

Subtracting 1 both sides

`a_q/a_p-1=a_r/a_q-1`

`(a_q-a_p)/a_p=(a_r-a_q)/a_q`

`(a_q-a_p)/(a_r-a_q)=a_p/a_q`

Putting value of `a_q,a_p,a_r`

`((a+(q-1)d)-(a+(p-1)d))/((a+(r-1)d)-(a+(q-1)d))=a_p/a_q`

`(d(q-1-p+1))/(d(r-1-q+1))=a_p/a_q`

`(q-1-p+1)/(r-1-q+1)=a_p/a_q`

`(q-p)/(r-q)=a_p/a_q`

`(r-q)/(q-p)=a_q/a_p`

`(-q-r)/-(p-q)=a_q/a_p`

`(q-r)/(p-q)=a_q/a_p " " " " ....(A)`


Taking `a_r/a_q=a_s/a_r`

Subtracting 1 both sides

`a_r/a_q-1=a_s/a_r-1`

`(a_r-a_q)/a_q=(a_s-a_r)/a_r`

`(a_s-a_r)/(a_r-a_q)=a_r/a_q`

`(a_r-a_s)/(a_q-a_r)=a_r/a_q`

Putting value of `a_q,a_s,a_r`

`((a+(r-1)d)-(a+(s-1)d))/((a+(q-1)d)-(a+(r-1)d))=a_r/a_q`

`(d(r-1-s+1))/(d(q-1-r+1))=a_r/a_q`

`(r-1-s+1)/(q-1-r+1)=a_r/a_q`

`(r-s)/(q-r)=a_r/a_q`

From `(1)`

`(r-s)/(q-r)=a_q/a_p " " " " ....(B)`


From `(A)` and `(B)`

`(q-r)/(p-q)=(r-s)/(q-r)`

Hence `(p-q),(q-r),(r-s)` are in GP

Hence Proved