Mathematics - Sequence & Series Question with Solution | TestHub

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

And the last term $=199=L$

Let the three term be $a,a+d,a+2d$

$a_n=L=a+\left(n-1\right)d$

$d_i=\frac{L-a}{n-1}$

$n\to$ number of terms

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

For, $n=3, d_1=\frac{199-100}{2}$

$=\frac{99}{2}\notin \mathrm{I}$

For, $n=4,{ d}_2=\frac{99}{3}=33\in I$

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For, $n=10,{ d}_3=\frac{99}{9}=11\in I$

For, $n=12,{ d}_4=\frac{99}{11}=9\in I$

So, sum of all common difference that are integer will be $\sum d_i=33+11+9=53$