If the 19^{th} term of a non-zero A.P. is zer | vedclass

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(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $t_n = a + (n - 1)d$.Given tha

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$3 : 1$

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(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $t_n = a + (n - 1)d$.Given that the $19^{th}$ term is zero,we have $t_{19} = a + 18d = 0$,which implies $a = -18d$.We need to find the ratio of the $49^{th}$ term to the $29^{th}$ term:$\frac{t_{49}}{t_{29}} = \frac{a + 48d}{a + 28d}$.Substituting $a = -18d$ into the expression:$\frac{t_{49}}{t_{29}} = \frac{-18d + 48d}{-18d + 28d} = \frac{30d}{10d} = 3$.Thus,the ratio is $3 : 1$.

If the $19^{th}$ term of a non-zero $A.P.$ is zero,then the ratio of its ($49^{th}$ term) to ($29^{th}$ term) is:

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