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find the smallest number of terms which may be taken in order that the sum of the arithmetical series 325+350+375+....may be exceed10000
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n/2{2*325+(n-1)25}= n/2{650+25n-25}=10000 n/2{625+25n}=10000 625n+25n^2=20000 n^2+25n=800 n=18.5(approximately) sum upto 19th term exceeds 10000
i do this but due my silly mistake, answer no match @Praja_01
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