Question

If the pth, qth , rth terms of the GP are a, b, c respectively. Prove that
${a}^{q - r} . {b}^{r - p} . {c}^{p - q} = 1$
​

Answer 1

${ \red{ \sf{given:-}}}$

${ \purple{ \sf{ a_{p} = a}}}$

${ \purple{ \sf{ a_{q} = b}}}$

${ \purple{ \sf{ a_{r} = c}}}$

L.H.S. = ${ \blue{ \sf{ {a}^{q - r} \: . \: {b}^{r - p} \: . \: {c}^{p - q}}}}$

Let,

${ \orange{ \sf{ {ar}^{p - 1} = a}}}$

${ \orange{ \sf{ {ar}^{q - 1}} = b}}$

${ \orange{ \sf{ {ar}^{r - 1} = c}}}$

By considering L.H.S.

${ \red{ \sf{( { {ar}^{p - 1}) }^{q - r} }}} \: . \: { \red{ \sf{( { {ar}^{q - 1}) }^{r - p} }}} \: . \: { \red{ \sf{( { {ar}^{r - 1}) }^{p - q} }}}$

${ \purple{ \sf{ {(a)}^{ \cancel{q} - \cancel{r} + \cancel{r} - \cancel{p} + \cancel{p} - \cancel{q}}}}} \: . \: { \purple{ \sf{ {(r)}^{(p - 1)(q - r)(q - 1)(r - p)(r - 1)(p - q)}}}}$

${ = \purple{ \sf{ {(a)}^{0} \: . \: { \purple{ \sf{ {(r)}^{ \cancel{pq} - \cancel{pq} - \cancel{q} + \cancel{r} + \cancel{qr} - \cancel{qp} - \cancel{r} + \cancel{p} + \cancel{qp} - \cancel{qr} - \cancel{p} + \cancel{q}}}}}}}}$

${ = \purple{ \sf {(a)}^{0} \: . \: {(r)}^{0}}}$

${ = \purple{ \sf{(1) \: . \: (1)}}}$

${ = \boxed{ \red{}{ \sf{1}}}}$

Answer 2

Answer:

Proof below.

Step-by-step explanation:

General form of a geometric sequence:

$\boxed{a_n=ar^{n-1}}$

Where:

• $a_n$ is the nth term.
• a is the first term.
• r is the common ratio.
• n is the position of the term.

If the pth, qth, rth terms of the geometric progression are a, b, c respectively, then:

$a_p=ar^{p-1}=a$

$a_q=ar^{q-1}=b$

$a_r=ar^{r-1}=c$

Substitute the expressions for a, b and c into the LHS of the given equation and solve:

$\large\begin{aligned} a^{q-r}\cdot b^{r-p}\cdot c^{p-q}&=(ar^{p-1})^{q-r}\cdot(ar^{q-1})^{r-p} \cdot (ar^{r-1})^{p-q}\\\\&=a^{(q-r)}\cdot r^{(p-1)(q-r)}\cdot a^{(r-p)} \cdot r^{(q-1)(r-p)} \cdot a^{(p-q)} \cdot r^{(r-1)(p-q)}\\\\&=a^{(q-r)}\cdot a^{(r-p)} \cdot a^{(p-q)} \cdot r^{(p-1)(q-r)}\cdot r^{(q-1)(r-p)}\cdot r^{(r-1)(p-q)}\\\\&=a^{(q-r)+(r-p)+(p-q)}\cdot r^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}\\\\&=a^{0}\cdot r^{(pq-pr-q+r)+(rq-pq-r+p)+(pr-rq-p+q)}\\\\&=a^0\cdot r^{0}\\\\&=1\cdot 1\\\\&=1\end{aligned}$

Hence proving that:

$a^{q-r}\cdot b^{r-p}\cdot c^{p-q}=1$

------------------------------------------------------------------------

Exponent rules used:

$(a^b)^c=a^{bc}$

$a^b \cdot a^c=a^{b+c}$

$a^0=1$