Question

If the $p^{th},q^{th},r^{th}$ and $s^{th}$ terms of an A.P. are in G.P,. then $p-q,q-r,r-s$ are in

• A. A.P
• B. G.P
• C. H.P
• D. None of the above

Answer 1

The correct option is B G.P

Let $p^{th}$ terms $=a+(p-1)d$,

$q^{th}$ terms $=a+(q-1)d$,

$r^{th}$ terms $=a+(r-1)d$,

$s^{th}$ terms $=a+(s-1)d$,

$\therefore$ Let $x$ and $y$ be the 1st term and common ratio of the G.P (As given) receptively.

So, $a+(p-1)d=x$ ...(1)

$a+(q-1)d=xy$ ...(2)

$a+(r-1)d=x{y}^2$ ...(3)

$a+(s-1)d=x{y}^3$ ...(4)

Apply (1) - (2)

$a+pd-d-a-q-qd+d=x-xy$

$(p-q)d=x(1-y)$ ...(5)

Apply (2)-(3), we get

$(q-r)d=xy(1-y)$ ...(6)

Apply (3) - (4)

$(r-s)d=x{y}^2(1-y)$ ...(7)

Now, (6) $÷$ (5)

$\frac{(q-r)d}{(p-q)d}=\frac{xy(1-y)}{x(1-y)}$

$\left(\frac{q-r}{p-q}\right)=y$ ...(8)

And, (7) $÷$ (6}

$\frac{(r-s)d}{(q-r)d}=\frac{x{y}^2(1-y)}{xy(1-y)}$

$\left(\frac{r-3}{q-2}\right)=y$ ...(9)

from, (8) and (9)

we get,

$\frac{(q-r)}{(p-q)}=\frac{(r-s)}{(q-r)}$

$\therefore$ Hence,

$(p-q),(q-r),(r-s)$ are in G.P.