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 $S_n = \frac{n}{2}[2a + (n - 1)d]$.Accord

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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 $S_n = \frac{n}{2}[2a + (n - 1)d]$.According to the problem,$S_{10} = 4 	imes S_5$.Substituting the formula:$\frac{10}{2}[2a + (10 - 1)d] = 4 	imes \frac{5}{2}[2a + (5 - 1)d]$$5[2a + 9d] = 4 	imes 2.5[2a + 4d]$$5[2a + 9d] = 10[2a + 4d]$Dividing both sides by $5$:$2a + 9d = 2[2a + 4d]$$2a + 9d = 4a + 8d$Rearranging the terms to find the ratio $a:d$:$9d - 8d = 4a - 2a$$d = 2a$$\frac{a}{d} = \frac{1}{2}$Therefore,the ratio of the first term to the common difference is $1:2$.

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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