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

Proof below.

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`

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Exponent rules used:

`(a^b)^c=a^(bc)`

`a^b \cdot a^c=a^(b+c)`

`a^0=1`