Different A.P. are constructed with the first term 100 , the last term 199, and integral common differences. The sum of the common differences of all such A.P. having at least 3 terms and at most 33 terms is : Sequence & Series (Mathematics)

Let the common differences of the A.P.'s having $3,4, \ldots, 33$ terms be $d_1, d_2, \ldots, d_{31}$ .So, $d_1=\frac{199-100}{2}=49.5 otin I, d_2=\frac{199-100}{3}=33 \in I$ , $d_3=\frac{199-100}{4}=24.75 otin I, d_4=\frac{199-100}{5}=19.8 otin I$ Continuing in this way, we get only 3 possibilities of $d$ i.e., $9,11,33$ .So, required sum $=9+11+33=53$

Explanation

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