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

English

Gujarati

Hindi

8. Sequences and Series

hard

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

Share

Similar Questions

Let $3,6,9,12, \ldots$ upto $78$ terms and $5,9,13,17, \ldots$ upto $59$ terms be two series. Then, the sum of the terms common to both the series is equal to

easy

(JEE MAIN-2022)

If the sum of $n$ terms of an $A.P.$ is $nA + {n^2}B$, where $A,B$ are constants, then its common difference will be

medium

The sides of a right angled triangle are in arithmetic progression. If the triangle has area $24$ , then what is the length of its smallest side ?

medium

(IIT-2017)

Three number are in $A.P.$ such that their sum is $18$ and sum of their squares is $158$. The greatest number among them is

easy

If $< {a_n} >$ is an $A.P$. and $a_1 + a_4 + a_7 + …….+ a_{16} = 147$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

normal