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JEE Mains · Maths · STD 11 - 8. sequence and series

The sum of all natural numbers \(‘n’\) such that \(100 < n < 200\) and \(H.C. F\, (91, n) > 1\) is

  1. A \(3221\)
  2. B \(3303\)
  3. C \(3203\)
  4. D \(3121\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(3121\)

Step-by-step Solution

Detailed explanation

\({S_A} = \) sum of numbers between \(100\) and \(200\) which are divisible by \(7\). \( \Rightarrow {S_A} = 105 + 112 + ..... + 196\) \({S_A} = \frac{{14}}{2}\left[ {105 + 196} \right] = 2107\) \({S_B} = \) Sum of number between \(100\) and \(200\) which are divisible by…
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