Different A.P.'s are constructed with the first term 100 , the last...

Given ${1}^{st}$ term of the A.P$= 100 = a$

And the last term $= 199 = L$

Let the three term be $a, a + d, a + 2d$

${a}_{n} = L = a + \left(n - 1\right)d$

${d}_{i} = \dfrac{L - a}{n - 1}$

$n \to$ number of terms

Now checking for what value of $n$ we have integral common difference,

For, $n = 3, {d}_{1} = \dfrac{199 - 100}{2}$

$= \dfrac{99}{2} \notin I$

For, $n = 4, {d}_{2} = \dfrac{99}{3} = 33 \in I$

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For, $n = 10, {d}_{3} = \dfrac{99}{9} = 11 \in I$

For, $n = 12, {d}_{4} = \dfrac{99}{11} = 9 \in I$

So, sum of all common difference that are integer will be $∑ {d}_{i} = 33 + 11 + 9 = 53$