if there are m A.P.'s beginning with unity whose common difference are 1,2,3.....m respectively, show that the sum of their nth terms is 1/2m(mn-m+n+1).

Asked by aashu.60 | 24th Aug, 2015, 09:01: AM

Expert Answer:

C o n s i d e r space a n space A P space w i t h space 1 space a s space t h e space i n i t i a l space t e r m space a n d space c o m m o n space d i f f e r e n c e space 1. T h u s space A P subscript 1 equals 1 comma 2 comma 3 comma 4 comma... therefore n t h space t e r m space o f space A P subscript 1 equals 2 plus open parentheses n minus 1 close parentheses cross times 1 equals 2 plus open parentheses n minus 1 close parentheses  C o n s i d e r space a n space A P space w i t h space 1 space a s space t h e space i n i t i a l space t e r m space a n d space c o m m o n space d i f f e r e n c e space 2. T h u s space A P subscript 2 equals 1 comma 3 comma 5 comma 7 comma... therefore n t h space t e r m space o f space A P subscript 2 equals 2 plus open parentheses n minus 1 close parentheses cross times 2 equals 2 plus open parentheses n minus 1 close parentheses 2  C o n s i d e r space a n space A P space w i t h space 1 space a s space t h e space i n i t i a l space t e r m space a n d space c o m m o n space d i f f e r e n c e space 3. T h u s space A P subscript 3 equals 1 comma 4 comma 7 comma 10 comma... therefore n t h space t e r m space o f space A P subscript 3 equals 2 plus open parentheses n minus 1 close parentheses cross times 3 equals 2 plus open parentheses n minus 1 close parentheses 3  . . . . C o n s i d e r space a n space A P space w i t h space 1 space a s space t h e space i n i t i a l space t e r m space a n d space c o m m o n space d i f f e r e n c e space m. T h u s space A P subscript 3 equals 1 comma m plus 1 comma 2 m plus 1 comma 3 m plus 1 comma... therefore n t h space t e r m space o f space A P subscript m equals 2 plus open parentheses n minus 1 close parentheses cross times m equals 2 plus open parentheses n minus 1 close parentheses m T h u s comma space s u m space o f space n t h space t e r m s space o f space a l l space t h e space A P s equals 1 plus open parentheses n minus 1 close parentheses plus 2 plus open parentheses n minus 1 close parentheses 2 plus 2 plus open parentheses n minus 1 close parentheses 3 plus... plus 2 plus open parentheses n minus 1 close parentheses m equals m plus open parentheses n minus 1 close parentheses open square brackets 1 plus 2 plus 3 plus... plus m close square brackets equals m plus open parentheses n minus 1 close parentheses open square brackets fraction numerator m open parentheses m plus 1 close parentheses over denominator 2 end fraction close square brackets equals m open square brackets 1 plus fraction numerator open parentheses n minus 1 close parentheses open parentheses m plus 1 close parentheses over denominator 2 end fraction close square brackets equals m open square brackets fraction numerator 2 plus m n minus m plus n minus 1 over denominator 2 end fraction close square brackets equals m open square brackets fraction numerator m n minus m plus n plus 1 over denominator 2 end fraction close square brackets

Answered by Vimala Ramamurthy | 24th Aug, 2015, 11:24: AM