If the sum of the 10 terms of an A.P. is 4 ti | vedclass

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(A) Let the first term be $a$ and the common difference be $d$. The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Given t

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$1:2$

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(A) Let the first term be $a$ and the common difference be $d$. The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Given that $S_{10} = 4 	imes S_5$,we have:$\frac{10}{2}[2a + (10-1)d] = 4 	imes \frac{5}{2}[2a + (5-1)d]$$5(2a + 9d) = 4 	imes 5(2a + 4d)$Dividing both sides by $5$:$2a + 9d = 4(2a + 4d)$$2a + 9d = 8a + 16d$$9d - 16d = 8a - 2a$$-7d = 6a$$\frac{a}{d} = -\frac{7}{6}$.Wait,re-evaluating the calculation: $2a + 9d = 8a + 16d$ leads to $6a = -7d$,which is not in the options. Let us re-check the equation: $S_{10} = 4S_5$ $\Rightarrow 5(2a+9d) = 4 	imes 2.5(2a+4d)$ $\Rightarrow 5(2a+9d) = 10(2a+4d)$ $\Rightarrow 2a+9d = 2(2a+4d)$ $\Rightarrow 2a+9d = 4a+8d$ $\Rightarrow d = 2a$ $\Rightarrow \frac{a}{d} = \frac{1}{2}$.

If the sum of the $10$ terms of an $A.P.$ is $4$ times the sum of its $5$ terms,then the ratio of the first term to the common difference is:

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