How many terms of a a.p 16 14 12 are needed to give a sum 60?explain how we got 2 answer
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Answer:
16,14,12
sum=n/2{2a+(n-1)d}
60=n/2{2*16+(n-1)(-2)}
120=n{34-2n}
120=34n-2n2
n2 -17n+60=0
n2 -12n-5n+60=0
n(n-12)-5(n-12)=0
n=5,12
here,the common difference is negative therefore terms go on diminishing and 9th term becomes zero.All terms after 9th term are negative.These negative terms when added to positive terms from 6th to 8th term,they cancel out each other and the sum remains same.
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