JEE PYQ: Sequence And Series Question 66
Question 66 - 2019 (11 Jan Shift 2)
If 19th term of a non-zero A.P. is zero, then its (49th term) : (29th term) is:
(1) 4:1
(2) 1:3
(3) 3:1
(4) 2:1
Show Answer
Answer: (3)
Solution
Let first term and common difference of AP be $a$ and $d$ respectively, then
$\therefore t_{19} = a + 18d = 0$
$\therefore a = -18d$
$\therefore \frac{t_{49}}{t_{29}} = \frac{a + 48d}{a + 28d} = \frac{-18d + 48d}{-18d + 28d} = \frac{30d}{10d} = 3$
$t_{49} : t_{29} = 3 : 1$
BETA
