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Worked Examples · Example 18

Q.If the mmth and nnth terms of a G.P are nn and mm respectively, show that its (m+n)(m+n)th term is (nmmn)1m−n\left(\dfrac{n^m}{m^n}\right)^{\frac{1}{m-n}}.

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Dividing the equations for the mmth and nnth terms isolates the common ratio rr; substituting back into am+n=an⋅rma_{m+n}=a_{n}\cdot r^{m} yields exactly the required expression.

nnth term of a G.P. with first term aa and common ratio rr:

ak=a rk−1a_k = a\,r^{k-1}

Given: am=na_m = n and an=ma_n = m (note the terms and their indices are swapped between the two conditions).

  1. Write the given conditions: am=a rm−1=na_m = a\,r^{m-1} = n … (i), and an=a rn−1=ma_n = a\,r^{n-1} = m … (ii).
  2. Divide (i) by (ii): a rm−1a rn−1=nm⇒rm−n=nm\dfrac{a\,r^{m-1}}{a\,r^{n-1}} = \dfrac{n}{m} \Rightarrow r^{m-n} = \dfrac{n}{m}.
  3. Solve for rr: r=(nm)1m−nr = \left(\dfrac{n}{m}\right)^{\frac{1}{m-n}}.
  4. From (ii), a=mrn−1a = \dfrac{m}{r^{n-1}}, i.e. a rn−1=ma\,r^{n-1} = m exactly (this is just equation (ii) restated).
  5. Now find the (m+n)(m+n)th term: am+n=a rm+n−1=(a rn−1)⋅rma_{m+n} = a\,r^{m+n-1} = \big(a\,r^{n-1}\big)\cdot r^{m}.
  6. Substitute a rn−1=ma\,r^{n-1}=m from (ii): am+n=m⋅rma_{m+n} = m\cdot r^m.
  7. Substitute r=(nm)1m−nr=\left(\dfrac{n}{m}\right)^{\frac{1}{m-n}} from step 3: am+n=m⋅(nm)mm−na_{m+n} = m\cdot\left(\dfrac{n}{m}\right)^{\frac{m}{m-n}}.
  8. Rewrite: m⋅nm/(m−n)mm/(m−n)=nmm−n⋅m1−mm−n=nmm−n⋅m(m−n)−mm−n=nmm−n⋅m−nm−nm\cdot\dfrac{n^{m/(m-n)}}{m^{m/(m-n)}} = n^{\frac{m}{m-n}}\cdot m^{1-\frac{m}{m-n}} = n^{\frac{m}{m-n}}\cdot m^{\frac{(m-n)-m}{m-n}} = n^{\frac{m}{m-n}}\cdot m^{-\frac{n}{m-n}}. …

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