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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


User Droebi
by
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2 Answers

18 votes
18 votes


{ \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}}}}

User Best
by
3.1k points
13 votes
13 votes

Answer:

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

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

Exponent rules used:


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


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


a^0=1

User Wasif Khan
by
3.0k points