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 given by $T_n = a + (n-1)d$.According to the first condition: $T_

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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 given by $T_n = a + (n-1)d$.According to the first condition: $T_9 = 5 	imes T_2$.$a + 8d = 5(a + d) \implies a + 8d = 5a + 5d \implies 4a = 3d \implies d = \frac{4a}{3}$.According to the second condition: $T_{13} = 2 	imes T_6 + 5$.$a + 12d = 2(a + 5d) + 5 \implies a + 12d = 2a + 10d + 5 \implies 2d - a = 5$.Substitute $d = \frac{4a}{3}$ into the equation $2d - a = 5$:$2(\frac{4a}{3}) - a = 5 \implies \frac{8a}{3} - a = 5 \implies \frac{5a}{3} = 5 \implies a = 3$.

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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