If frac{3 + 5 + 7 + dots + (2n + 1)}{5 + 8 + | vedclass

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(A) The sum of $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.For the numerator: $a = 3, d = 2$. Sum $= \frac{n

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$35$

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(A) The sum of $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.For the numerator: $a = 3, d = 2$. Sum $= \frac{n}{2}[6 + (n - 1)2] = \frac{n}{2}[2n + 4] = n(n + 2)$.For the denominator: $10$ terms,$a = 5, d = 3$. Sum $= \frac{10}{2}[2(5) + (10 - 1)3] = 5[10 + 27] = 5 	imes 37 = 185$.Given $\frac{n(n + 2)}{185} = 7$.$n^2 + 2n = 1295$.$n^2 + 2n - 1295 = 0$.$(n + 37)(n - 35) = 0$.Since $n$ must be positive,$n = 35$.

If $\frac{3 + 5 + 7 + \dots + (2n + 1)}{5 + 8 + 11 + \dots + (3n + 2)} = 7$,find the value of $n$. (Note: The original problem implies $n$ terms in the numerator and $10$ terms in the denominator as per the provided solution structure).

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