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How many terms of the A.P. 20, 19 1/3, 1...

How many terms of the A.P. `20, 19 1/3, 18 2/3,`……… must be taken so that their sum is 300?

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Verified by Experts

Here, given series is an AP with first term `a=20` and the common difference, `d=-(2)/(3).`
Let the sum of n terms of the series be 300.
Then, `S_(n)=(n)/(2){2a+(n-1)d}`
`implies 300=(n)/(2){2xx20+(n-1)(-(2)/(3))}`
` implies 300=(n)/(3){60=n+1}`
`implies n^(2)-61n+900=0` `implies (n-25)(n=36)=0`
`implies n=25 " or " n=36`
`therefore` Sum of 25 terms = Sum of terms = 300
Explaination of double answer
Her, the common difference is negetive, therefore terms go on diminishing and `t_(31)=20+(31-1)((-2)/(3))=0 i.e.,31st` term becomes zero. all terms after 31st term are negative. These negative terms `(t_(32),t_(33),t_(34),t_(35)t_(36))` when added to positive terms `(t_(26),t_(27),t_(28),t_(29)t_(30))`, they cancel out each other i.e., sum of ter,s from 26th to 36th terms is zero. Hence, the sum of 25 terms as well as that of 36 terms is 300.
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