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If the `p t h , q t h a n d r t h`terms of `a`G.P. are `a , b , c`respectively, prove that: `a^((q-r))dot^( )b^((r-p))dotc^((p-q))=1.` |
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Answer» Let A , D are the first term and common difference difference of Ap and x, R are the first term and common ratio fo Gp respectively . According to the given condition , `A+(p-1)d=2`…(i) `A+(q-1) d=b`***(iii) ` A +(r-1) d=c`...(iii) and `a=xR^(p-1)` ...(iv) `b=xR^(p-1)` ...(v) `c=xR^(p-1)` ***(vi) On subtracting Eq,(ii) from Eq (i) we get `d(p-1-q+1)=a-b` `a-b=d(p-q)`***(vii) On subtracting Eq,(iii) from Eq (ii) we get `d(p-1-r+1)=b-c` `implies b-c=d(q-r)` ...(viii) On subtracting Eq,(i) from Eq (iii) we get `d(r-1-p+1)=c-a` `c-a=d (r-p)` ....(ix) taking LHS `=a^(b-c) b^(c-a) c^(a-b)` using Eqs (iv) ,(v) ,(vi) and (viii) , (ix) `LHS =(xR^(p-1))^(d(q-r) )(xR^(q-1) ) ^(d(r-p) ) (xR^(r-1))^(d(p-q))` `=x^(d(q-1)+d(r-p)+d(p-q))R^((p-1)d(q-r)+d(r-p)+(r-1)d(p-q))` `=x^(d(q-r+r=p-p-q))` `R^(d(pq-pr-q+r+qr- pq -r +p rp -rp-p+q))=x^(0)r^(0)=1` =RHS Hence proved . |
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