If the sum of the first 10 terms of an arithm | vedclass

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

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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 the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Given that $S_{10} = 4 	imes S_5$.Substituting the formula:$\frac{10}{2}[2a + 9d] = 4 	imes \frac{5}{2}[2a + 4d]$$5(2a + 9d) = 10(2a + 4d)$$10a + 45d = 20a + 40d$$45d - 40d = 20a - 10a$$5d = 10a$$\frac{a}{d} = \frac{5}{10} = \frac{1}{2}$Therefore,the ratio $a:d = 1:2$.

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

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