In the solution of Most important questions maths I have a doubt in the third question.
In a step,we replace n by 2m-1.I cant understand how do we derive that.
Please help.

Asked by SHUBHAM GHODE | 14th Mar, 2017, 07:45: PM

Expert Answer:

DEAR STUDENT, MY SINCERE APOLIGIES. This is regarding the APPLICATION OF TRIGONOMETRY, 3RD QUESTION.
 
The NO is missing. There is NO SLACK. I sent your doubt in this query so that you do not lose you chance to ask doubts. Regret the inconvenience. We shall get that rectified. Thank you Shubam. The first diagram that you drew is right.
 
 
 
begin mathsize 16px style Let space straight S subscript straight A space and space straight S subscript straight B space be space the space sums space straight n space terms space of space two space APs.
Let space straight a subscript straight A space and space straight a subscript straight B space be space the space initial space terms space and space straight d subscript straight A space and space straight d subscript straight B space be space the space common space differences space of space two space APs.
Given space that space the space ratio space of space the space sum space of space two space APs space is space open parentheses 7 straight n plus 1 close parentheses colon open parentheses 4 straight n plus 27 close parentheses
rightwards double arrow straight S subscript straight A over straight S subscript straight B equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction
rightwards double arrow fraction numerator begin display style straight n over 2 open square brackets 2 straight a subscript straight A plus open parentheses straight n minus 1 close parentheses straight d subscript straight A close square brackets end style over denominator straight n over 2 open square brackets 2 straight a subscript straight B plus open parentheses straight n minus 1 close parentheses straight d subscript straight B close square brackets end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction
rightwards double arrow fraction numerator begin display style 2 straight a subscript straight A plus open parentheses straight n minus 1 close parentheses straight d subscript straight A end style over denominator 2 straight a subscript straight B plus open parentheses straight n minus 1 close parentheses straight d subscript straight B end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction
rightwards double arrow fraction numerator begin display style fraction numerator begin display style 2 straight a subscript straight A plus open parentheses straight n minus 1 close parentheses straight d subscript straight A end style over denominator 2 end fraction end style over denominator begin display style fraction numerator 2 straight a subscript straight B plus open parentheses straight n minus 1 close parentheses straight d subscript straight B over denominator 2 end fraction end style end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction
rightwards double arrow fraction numerator begin display style straight a subscript straight A plus fraction numerator begin display style open parentheses straight n minus 1 close parentheses straight d subscript straight A end style over denominator 2 end fraction end style over denominator begin display style straight a subscript straight B plus fraction numerator open parentheses straight n minus 1 close parentheses straight d subscript straight B over denominator 2 end fraction end style end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction
rightwards double arrow fraction numerator begin display style straight a subscript straight A plus fraction numerator begin display style open parentheses straight n minus 1 close parentheses end style over denominator 2 end fraction straight d subscript straight A end style over denominator begin display style straight a subscript straight B plus fraction numerator open parentheses straight n minus 1 close parentheses over denominator 2 end fraction straight d subscript straight B end style end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction... left parenthesis 1 right parenthesis
Assume space fraction numerator open parentheses straight n minus 1 close parentheses over denominator 2 end fraction equals open parentheses straight m minus 1 close parentheses
rightwards double arrow straight n minus 1 equals 2 open parentheses straight m minus 1 close parentheses
rightwards double arrow straight n minus 1 equals 2 straight m minus 2
rightwards double arrow straight n equals 2 straight m minus 2 plus 1
rightwards double arrow straight n equals 2 straight m minus 1
Thus comma space substituting space the space values comma space fraction numerator open parentheses straight n minus 1 close parentheses over denominator 2 end fraction equals open parentheses straight m minus 1 close parentheses space and space straight n equals 2 straight m minus 1 space in space equation space left parenthesis 1 right parenthesis comma space we space have comma
fraction numerator begin display style straight a subscript straight A plus open parentheses straight m minus 1 close parentheses straight d subscript straight A end style over denominator begin display style straight a subscript straight B plus open parentheses straight m minus 1 close parentheses straight d subscript straight B end style end fraction equals fraction numerator 7 open parentheses 2 straight m minus 1 close parentheses plus 1 over denominator 4 open parentheses 2 straight m minus 1 close parentheses plus 27 end fraction
rightwards double arrow straight t subscript mA over straight t subscript mB equals fraction numerator 14 straight m minus 7 plus 1 over denominator 8 straight m minus 4 plus 27 end fraction
rightwards double arrow straight t subscript mA over straight t subscript mB equals fraction numerator 14 straight m minus 6 over denominator 8 straight m plus 23 end fraction
Thus space the space ratio space of space straight m space terms space of space two space APs space is space 14 straight m minus 6 colon 8 straight m plus 23
end style
 
 
begin mathsize 16px style Dear space student comma space the space above space is space the space solution. space Below space is space the space answer space to space your space question. space
See space in space the space question comma space we space have space to space find space theratio space of space the space mth space term space of space the space two space APs.
mth space term equals bold t subscript bold m bold equals bold a bold plus bold left parenthesis bold m bold minus bold 1 bold right parenthesis bold d
If space you space observe space the space equation comma space left parenthesis 1 right parenthesis comma space you space can space see space that space with space the space help space of space the space above space bolded space formula comma
if space we space try space to space get space an space equation space of space the space type space where space we space have apostrophe straight m apostrophe space instead space of space straight n comma space we space can space acheive space this.
We space can space get space this space only space if space we space replace space fraction numerator straight n minus 1 over denominator 2 end fraction space by space straight m minus 1 space and space hence comma
we space will space get space straight n minus 1 equals 2 straight m minus 2 rightwards double arrow straight n equals 2 straight m minus 1 space which space is space what space we space want.

Try space putting space the space same comma space that space is comma space fraction numerator straight n minus 1 over denominator 2 end fraction equals straight m
When space yousolve space further comma space you space will space get space fraction numerator straight a subscript straight A plus md subscript straight A over denominator straight a subscript straight B plus md subscript straight B end fraction equals fraction numerator 7 left parenthesis 2 straight m plus 1 right parenthesis plus 1 over denominator 4 left parenthesis 2 straight m plus 1 right parenthesis plus 27 end fraction
In space this space case space you space cannot space use space the space bolded space formula left parenthesis bold t subscript bold m comma space term space formula right parenthesis.
Hence comma space we space substitute space straight n equals 2 straight m minus 1. end style
Hope this clears your doubt.

Answered by Rebecca Fernandes | 31st Mar, 2017, 09:40: PM

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