If 19^{th} terms of non -zero A.P. is zero, then its ( 49^{th} term) : ( 29^{th} term) is

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8. Sequences and Series

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

A

$4 : 1$

B

$1 : 3$

C

$3 : 1$

D

$2 : 1$

(JEE MAIN-2019)

Solution

$a + 18d = 0 \Rightarrow a = – 18d$

$\frac{{{t_{49}}}}{{{t_{29}}}} = \frac{{a + 48d}}{{a + 28d}} = \frac{{ – 18d + 48d}}{{ – 18d + 28d}}$

$ = \frac{{30d}}{{10d}} = 3$

Standard 11

Mathematics

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