Question

if pth, qth, rth term of an gp are a, b,c respectively prove that a q-r ,br-p ,cp-q =1​

Answer 1

EXPLANATION.

→ Pth, qth, rth term of an Gp are a, b, c

$\sf \: to \: prove \: = \: {a}^{q - r} {b}^{r - p} {c}^{p - q} = 1$

$\sf \:as \: we \: know \: that \\ \\ \sf \:nth \: terms \: of \: an \: gp = ar {}^{n - 1} \\ \\ \sf \:first \: term \: of \: an \: gp \: = a \\ \\ \sf \:common \: ratio \: = r$

$\sf \: {p}^{th} \: term \: of \: gp = a \\ \\ \sf \: \: {Ar}^{p - 1} = a \\ \\ \sf \: a \: = {Ar}^{p - 1} \\ \\ \sf : \implies \: {a}^{q - r} = ( {Ar}^{p - 1} ) {}^{q - r} .......(1)$

$\sf \: {q}^{th} \: term \: of \: gp \: = b \\ \\ \sf \: {Ar}^{q \: - 1} = b \\ \\ \sf \: b \: = {Ar}^{q \: - 1} \\ \\ \sf : \implies \: { b }^{r - p} = ( {Ar}^{q \: - 1} ) {}^{r - p} \: \: ........(2)$

$\sf \: {r}^{th} \: term \: of \: gp \: = c \\ \\ \sf \: {Ar}^{r \: - 1} = c \\ \\ \sf \: c \: = {Ar}^{r \: - 1} \\ \\ \sf : \implies \: {c}^{p - q} = ( {Ar}^{r \: - 1}) {}^{p - q} \: \: ......(3)$

$\sf: \implies \: put \: the \: value \: in \:equation \\ \\ \sf : \implies \: {a}^{q - r} {b}^{r - p} {c}^{p - q} \\ \\ \sf : \implies \: ({Ar}^{p \: - 1} ){}^{q - r} \times ( {Ar}^{q \: - 1}) {}^{r - p} \times ( {Ar}^{r \: - 1}) {}^{p - q} \\ \\ \sf : \implies \: A {}^{q - r} {r}^{(p - 1)(q - r)} \times A {}^{r - p} {r}^{(q - 1)(r - p)} \times A {}^{p - q} {r}^{(r - 1)(p - q)}$

$\sf : \implies \: A {}^{q - r} A {}^{r - p}A {}^{p - q} \times \: r {}^{(p - 1)(q - r)} {r}^{(q - 1)(r - p)} {r}^{(r - 1)(p - q)} \\ \\ \sf : \implies \: A {}^{(q - r) + (r - p) + (p - q)} \times \: {r}^{(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)} \\ \\ \sf : \implies \: A {}^{0} \times {r}^{(pq - pq - q + q + r - r + p - p + qr - qr + qp - qr)} \\ \\ \sf : \implies \: A {}^{0} \times {r}^{0} \\ \\ \sf : \implies \: 1 = proved$