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(A) The numerator is an arithmetic progression with first term $a_1 = 3$ and common difference $d_1 = 2$. The sum of $n$ terms is $S_n = \frac{n}{2}[2

Vedclass · Accepted Answer

$35$

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(A) The numerator is an arithmetic progression with first term $a_1 = 3$ and common difference $d_1 = 2$. The sum of $n$ terms is $S_n = \frac{n}{2}[2(3) + (n - 1)2] = \frac{n}{2}[6 + 2n - 2] = \frac{n}{2}[2n + 4] = n(n + 2)$.The denominator is an arithmetic progression with first term $a_2 = 5$ and common difference $d_2 = 3$. The sum of $10$ terms is $S_{10} = \frac{10}{2}[2(5) + (10 - 1)3] = 5[10 + 27] = 5 	imes 37 = 185$.Given the equation $\frac{n(n + 2)}{185} = 7$,we have $n^2 + 2n = 1295$.$n^2 + 2n - 1295 = 0$.Solving the quadratic equation using the formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $n = \frac{-2 \pm \sqrt{4 - 4(1)(-1295)}}{2} = \frac{-2 \pm \sqrt{5184}}{2} = \frac{-2 \pm 72}{2}$.Since $n$ must be positive,$n = \frac{70}{2} = 35$.

If $\frac{3 + 5 + 7 + \dots \text{ to } n \text{ terms}}{5 + 8 + 11 + \dots \text{ to } 10 \text{ terms}} = 7$,then the value of $n$ is

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