Different A.P.s are constructed with the firs | vedclass

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(D) Let the first term be $a = 100$ and the last term be $\ell = 199$. For an $A.P.$ with $n$ terms,the last term is given by $\ell = a + (n - 1)d$,wh

Vedclass · Accepted Answer

$53$

Vedclass · Answer

(D) Let the first term be $a = 100$ and the last term be $\ell = 199$. For an $A.P.$ with $n$ terms,the last term is given by $\ell = a + (n - 1)d$,where $d$ is the common difference. Thus,$d = \frac{\ell - a}{n - 1} = \frac{199 - 100}{n - 1} = \frac{99}{n - 1}$. We are given that $d$ must be an integer,so $(n - 1)$ must be a divisor of $99$. The divisors of $99$ are $1, 3, 9, 11, 33, 99$. We are given that the number of terms $n$ satisfies $3 \le n \le 33$. This implies $2 \le n - 1 \le 32$. The divisors of $99$ that satisfy this condition are $3, 9, 11$. For $n - 1 = 3$,$d = \frac{99}{3} = 33$. For $n - 1 = 9$,$d = \frac{99}{9} = 11$. For $n - 1 = 11$,$d = \frac{99}{11} = 9$. The sum of all such common differences is $33 + 11 + 9 = 53$.

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

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