(C) Let the first term 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 $19^{th}$

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(C) Let the first term 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: $t_{19} = a + 18d = 0$,which implies $a = -18d$.We need to find the ratio $\frac{t_{49}}{t_{29}}$.$t_{49} = a + 48d = -18d + 48d = 30d$.$t_{29} = a + 28d = -18d + 28d = 10d$.Therefore,the ratio is $\frac{30d}{10d} = \frac{3}{1}$,which is $3 : 1$.

Class 11 Mathematics Sequences and Series Arithmetic progression

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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:

JEE MAIN 2019 All JEE MAIN

A
$4 : 1$
B
$1 : 3$
C
$3 : 1$
D
$2 : 1$

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If the sum of $p$ terms of an arithmetic progression is equal to the sum of its $q$ terms,then what is the sum of its $(p + q)$ terms?

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Let $\sum_{k=1}^{n} a_{k} = \alpha n^{2} + \beta n$. If $a_{10} = 59$ and $a_{6} = 7a_{1}$, then $\alpha + \beta$ is equal to:

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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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