Explanation
no.\>of\>terms\>with\>same\>values\>are\>in\>AP(let's\>call\>them\>as\>a\>group)\\100=(\frac{n(n+1)}{2})\\n(n+1)=200\\for\>n\>=14\\n(n+1)=210\>>\>200\\hence\>all\>terms\>with\>values\>14,\>will\>cover\>group\>including\>100th\>term
given relation is \frac{1}{n+1} + \frac{1}{n+2}+.....+ \frac{1}{2n} > \frac{13}{24}< for all natural numbers n>1
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