Question

If pth, qth, rth terms of a G.P. are a, b, c respectively, then prove that value of a^q-r b^r-p c^p-q = 1



plz help me​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that first term of GP series is A and common ratio is R.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an geometric sequence GP is

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\: {r}^{n - 1} }}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.

According to statement,

It is given that,

$\rm :\longmapsto\:a_p = a$

$\bf\implies \: {AR}^{p - 1} = a - - - (1)$

Also,

$\rm :\longmapsto\:a_q = b$

$\bf\implies \: {AR}^{q - 1} = b - - - (2)$

Also,

$\rm :\longmapsto\:a_r = c$

$\bf\implies \: {AR}^{r - 1} = c - - - (3)$

Now, Consider

$\rm :\longmapsto\: {a}^{q - r} \: {b}^{r - p} \: {c}^{p - q}$

$\rm \: = \: \: {\bigg( {AR}^{p - 1} \bigg) }^{q - r} {\bigg( {AR}^{q - 1} \bigg) }^{r - p} {\bigg( {AR}^{r - 1} \bigg) }^{p - q}$

$\rm \: = \: {A}^{q - r} {A}^{r - p} {A}^{r - p} {\bigg( {R}^{p - 1} \bigg) }^{q - r} {\bigg( {R}^{q - 1} \bigg) }^{r - p} {\bigg( {R}^{r - 1} \bigg) }^{p - q}$

$\rm \: = \: {\bigg(A\bigg) }^{q - r + r - p + p - q} {\bigg(R\bigg) }^{(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)}$

$\rm \: = \: {\bigg(A\bigg) }^{0} {\bigg(R\bigg) }^{pq - pr - q + r + qr - qp - q + p + rp - rq - p + q}$

$\rm \: = \: \:1 \times {R}^{0}$

$\rm \: = \: \:1$

Hence, Proved

Additional Information :-

↝ Sum of n  terms of an Geometric sequence GP is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{a \: ( {r}^{n} \: - \: 1)}{r \: - \: 1 \:} \: provided \: that \: r \: \ne \: 1}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of GP.

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.