Sponsor Area

Permutations And Combinations

Question
CBSEENMA11012998

In an A.P., if the pth term is 1 over straight q and the qth term is 1 over straight p comma prove that the sum of the first pq terms must be 1 half left parenthesis pq space plus space 1 right parenthesis. where WiredFaculty

Solution

Let a be the first term and d be the common difference.
           WiredFaculty                                ...(i)
           straight t subscript straight q space equals space 1 over straight p space rightwards double arrow space space straight a space plus space left parenthesis straight q minus 1 right parenthesis straight d space equals space 1 over straight p                                     ...(ii)
Subtracting (ii) from (i), we get
space space left parenthesis straight p minus 1 right parenthesis straight d space minus space left parenthesis straight q minus 1 right parenthesis straight d space equals space 1 over straight q minus space 1 over straight p            or           left parenthesis straight p minus 1 space minus straight q plus 1 right parenthesis straight d space equals space fraction numerator straight p minus straight q over denominator pq end fraction
or             left parenthesis straight p minus straight q right parenthesis straight d space equals space fraction numerator straight p minus straight q over denominator pq end fraction                 or                             space space straight d equals 1 over pq
Using this value in (i), we get
    straight a plus left parenthesis straight p minus 1 right parenthesis 1 over pq space equals space 1 over straight q                         or               WiredFaculty
or            straight a plus 1 over straight q minus 1 over pq equals space 1 over straight q                 or                           straight a equals 1 over pq
                  WiredFaculty                               WiredFaculty
                        equals pq over 2 open square brackets 2.1 over pq plus left parenthesis pq minus 1 right parenthesis 1 over pq close square brackets space equals space pq over 2 open square brackets 2 over pq plus 1 minus 1 over pq close square brackets
                         space space equals pq over 2 open square brackets 1 over pq plus 1 close square brackets space equals space pq over 2 open square brackets fraction numerator 1 plus pq over denominator pq end fraction close square brackets space equals space 1 half left parenthesis pq plus 1 right parenthesis

Some More Questions From Permutations and Combinations Chapter

Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.