SATHEE: Sequences and Series 1 Question 2

Sequences and Series 1 Question 2

2. If 19th term of a non-zero AP is zero, then its (49th term) : (29th term) is

(2019 Main, 11 Jan II)

(a) $1 : 3$

(b) $4 : 1$

(d) $3 : 1$

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Answer:

Correct Answer: 2. (d)

Solution:

Let $t_{n}$ be the $n$ th term of given AP. Then, we have $t_{19} = 0$

$\begin{aligned} \Rightarrow a + (19 - 1) d & = 0 & [∵ t_{n} = a + (n - 1) d] \\ \Rightarrow a + 18 d & = 0 \\ Now, \frac{t_{49}}{t_{29}} & = \frac{a + 48 d}{a + 28 d} \\ = \frac{- 18 d + 48 d}{- 18 d + 28 d} [using Eq. (i)] \\ = \frac{30 d}{10 d} = 3 : 1 \end{aligned}$

Sequences and Series 1 Question 2

2. If 19th term of a non-zero AP is zero, then its (49th term) : (29th term) is

Answer:

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Sequences and Series 1 Question 2

2. If 19th term of a non-zero AP is zero, then its (49th term) : (29th term) is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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