Question

12. If S S S , and 1 2 3 are the sum of first n, 2n and 3n terms of a geometric series respectively,then prove that

Answer 1

Steps:
1) Let the GP be :
   $a,ar,a r^{2},....\\ [tex]S _{1}= \frac{a( r^{n}-1 )}{r-1}$

S₂  = $\frac{a( r^{2n}-1 )}{r-1} = \frac{a( r^{n}-1 )*( r^{n}+1 )}{r-1} =S _{n} ( r^{n}+1 )$
$S_{3n} = \frac{a( r^{3n}-1 )}{r-1} = \frac{a( r^{n}-1 )*( r^{2n}+ r^{n}+1 )} {r-1} = S _{n}* {( r^{2n}+ r^{n}+1 )}$



2)  Now,
     LHS = [tex] S_{1} ( S_{3}- S_{2} ) \\ =\ \textgreater \ S_{1}( S_{1}*( r^{2n}+r^{n}+1 )-S_{1} (r^{n}+1))\\ =\ \textgreater \ (S_{1})^{2}* {r^{2n}}\\=\ \textgreater \ (S_{1}*r^{n})^{2}\\ =\ \textgreater \ (S_{1}(r^{n}+1-1))^{2}\\=\ \textgreater \ (S_{1}(r^{n}+1)-S_{1})^{2}\\=\ \textgreater \ (S_{2}-S_{1})^{2}[/tex]


Hence Proved