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How many terms of the A.P. 16,14,12……….. are needed to given the sum `60` ? Explain why do we get two answers. |
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Answer» Here `S_(n)=60, a=16, d=-2, n=?` `S_(hn)=n/2[(2a+(n-1)d]`………..(Formula) `:.60=n/2[2xx16+(n-1)xx(-2)]` ….[Substitutig the values] `:.60=n/2(32-2n+2)` `:.60=n/2(34-2n)` `:.60=n/2xx2(17-n)` `:.60=n(17-n)` `:.60=17n-n^(2)` `:.n^(2)-17n+60=0` `:.n^(2)-5n-12n+60=0` `:.n(n-5)-12(n-5)=0` `:.(n-5)(n-12)=0` `:.n-5=0` or `n-12=0` `:.n=5` or `n=12` The required terms are 5 or 12. Explanation: THe common difference d of the A.P. is `-2` `:.` The terms of the A.P. are in descending order. Taking `n=5` the first 5 terms are 16,14,12,10,8. The sum is 60. Taking `n=12`, the last 7 terms `(12-5)` are `6,4,2,0,-2,-4,-5` The sum of these seven terms is 0. `:.` The sum of first 12 terms is also 60. The sum of the first terms `=` the sum of the first twelve terms. `:.` we get two answers. Ans. 5 terms or 12 terms. |
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