Answer :

Here, first term = a =20

Common difference = d = 58/3 - 20 = - 2/3


Let first n terms of the AP sums to 300.


Sn = 300


To find: n


Now, Sn = (n/2) × [2a + (n - 1)d]


Since, Sn = 300


(n/2) × [2a + (n - 1)d] = 300


(n/2) × [2(20) + (n - 1)(-2/3)] = 300


(n/2) × [40 - (2/3)n + (2/3)] = 300


(n/2) × [(120 - 2n + 2)/3] = 300


n[122 - 2n] = 1800


122n - 2n2 = 1800


2n2 - 122n + 1800 = 0


n2 - 61n + 900 = 0


(n - 36)( n - 25)= 0


n = 36 or n = 25


n= 36 or n = 25


Now, S36 = (36/2)[2a + 35d]


= 18(40 + 35(-2/3))


= 18(120 - 70)/3


= 6(50)


= 300


Also, S25 = (25/2)[2a + 24d]


= (25/2)(40 + 24(-2/3))


= (25/2)(40 - 16)


= (24 × 25)/2


= 12 × 25


= 300


Now, sum of 11 terms from 26th term to 36th term = S36 - S25 = 0


Both the first 25 terms and 36 terms give the sum 300 because the sum of last 11 terms is 0. So, the sum doesn’t get affected.


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