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) $4: 1$
(c) $2: 1$
(d) $3: 1$
Show Answer
Answer:
Correct Answer: 2. (d)
Solution:
- Let $t _n$ be the $n$th term of given AP. Then, we have $t _{19}=0$
$$ \begin{array}{rlr} \Rightarrow a+(19-1) d & =0 & {\left[\because t _n=a+(n-1) d\right]} \\ \Rightarrow \quad a+18 d & =0 \\ \text { Now, } \quad \frac{t _{49}}{t _{29}} & =\frac{a+48 d}{a+28 d} \\ & =\frac{-18 d+48 d}{-18 d+28 d} \quad \text { [using Eq. (i)] } \\ & =\frac{30 d}{10 d}=3: 1 \end{array} $$
