When the 9^{th} term of an A.P. is divided by | vedclass

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(B) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term is $T_n = a + (n-1)d$.Given $T_9 = 5T_2$ $\Rightarrow a + 8d = 5(a +

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$3$

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(B) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term is $T_n = a + (n-1)d$.Given $T_9 = 5T_2$ $\Rightarrow a + 8d = 5(a + d)$ $\Rightarrow a + 8d = 5a + 5d$ $\Rightarrow 4a - 3d = 0$ $\Rightarrow d = \frac{4a}{3}$.Given $T_{13} = 2T_6 + 5$ $\Rightarrow a + 12d = 2(a + 5d) + 5$ $\Rightarrow a + 12d = 2a + 10d + 5$ $\Rightarrow 2d - a = 5$.Substitute $d = \frac{4a}{3}$ into the second equation: $2(\frac{4a}{3}) - a = 5$.$\frac{8a}{3} - a = 5$ $\Rightarrow \frac{5a}{3} = 5$ $\Rightarrow a = 3$.

When the $9^{th}$ term of an $A.P.$ is divided by its $2^{nd}$ term,the quotient is $5$. When the $13^{th}$ term is divided by the $6^{th}$ term,the quotient is $2$ and the remainder is $5$. Find the first term of the $A.P.$

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