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Arithmetic Progressions

Question
CBSEENMA10007587

Given a3 = 75, S10 = 120, find ‘d’ and a10.

Solution

Here,  a= 15
         S10  = 120
We know that,
       straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket
therefore space space straight S subscript 10 equals 10 over 2 space left square bracket 2 straight a plus left parenthesis 10 minus 1 right parenthesis straight d right square bracket
rightwards double arrow space 120 space equals space 5 space straight x space left square bracket 2 straight a plus 9 straight d right square bracket
rightwards double arrow 2 straight a plus 9 straight d equals 120 over 5 equals 24
rightwards double arrow space 2 straight a plus 9 straight d space equals space 24 space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis
And. space space space straight a subscript 1 equals 15
rightwards double arrow space space straight a space plus 2 straight d space equals space 15 space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis
Solving (i) and (ii)
           5d = -6
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rightwards double arrow space space space straight a space equals space 15 plus 12 over 5
equals fraction numerator 75 plus 12 over denominator 5 end fraction equals 87 over 5
Now comma space space space space straight a subscript 10 equals straight a plus 9 straight d equals 87 over 5 plus 9 open parentheses fraction numerator negative 6 over denominator 5 end fraction close parentheses
space space space space space space space space space space space space space space space space space space space space equals 87 over 5 minus 54 over 5 equals fraction numerator 87 minus 54 over denominator 5 end fraction equals 33 over 5
Hence comma space space straight d space equals space fraction numerator negative 6 over denominator 5 end fraction space and space straight a subscript 10 space equals space 33 over 5