An AP consists of 12 term.the sum of the middle two term is 62.the sum of the last three term is 120.find the numbers
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Given that AP consists of 12 numbers.
As 12 is even..the middle two terms will be
(12/2) =6th term and {(12/2)+1}=7th term.
As pet question, 6th term + 7th term=62.....(1)
Therefore, the only possibility to satisfy above equation (1) is 6th term=30 and 7th term=32
it indicates that the common difference of AP is=2
Therefore the required numbers in AP are...
20,22,24,......,30,32,34,....38,40,42.
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