Permutations and Combinations

Permutations and Combinations

Question

Let the sum of n, 2n, 3n terms of an A.P., be S1, S2, S3, respectively, show that

straight S subscript 3 space equals space 3 left parenthesis straight S subscript 2 minus straight S subscript 1 right parenthesis

Answer

Let a be the first term and d be the common difference.

∴                        straight S subscript 1 space equals space straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket                       ...(i)
                 straight S subscript 2 space equals space fraction numerator 2 straight n over denominator 2 end fraction left square bracket 2 straight a plus left parenthesis 2 straight n minus 1 right parenthesis straight d right square bracket                            ...(ii)
                straight S subscript 3 space equals space fraction numerator 3 straight n over denominator 2 end fraction left square bracket 2 straight a plus left parenthesis 3 straight n minus 1 right parenthesis straight d right square bracket                             ... (iii)
Subtracting (i) from (ii), we get
         straight S subscript 2 minus straight S subscript 1 space equals space fraction numerator 2 straight n over denominator 2 end fraction left square bracket 2 straight a plus left parenthesis 2 straight n minus 1 right parenthesis straight d right square bracket space minus straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket
                     equals straight n over 2 left square bracket 4 straight a space plus space left parenthesis 4 straight n minus 2 right parenthesis straight d minus 2 straight a minus left parenthesis straight n minus 1 right parenthesis straight d right square bracket space equals space straight n over 2 left square bracket 2 straight a plus left parenthesis 3 straight n minus 1 right parenthesis straight d right square bracket
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#6 {main}</pre>
        

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