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IPMAT/Algebra
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its basically repetitions in the sequence of 1,2,3,... till 400 so sn>=400 so n is our 400th term
but how?
its basically 1 getting repeated once , 2 twice , 3 thrice and so on
Ignore what term is in the series treat each term as one object now if u look logically the series became 1 object+2 objects+3 objects and so on. Here we have to find a Sum of nth term of Ap(1,2,3...) which is greater than 400 as Sum of n-1 term is less than 400 so the formula will be (n(n+1))/2 > 400 now dont solve this instead put value by trying 2_3 times u will land on 28*29/2
Hence ans is 28