If 19 th term of a non-zero A.P. is zero, then its (49th term): (29th term) is:
If 19 th term of a non-zero A.P. is zero, then its (49th term):
(29th term) is:
4: 1
1: 3
3: 1
2: 1
Solution
Let first term and common difference of $A P$ be a and d respectively, then
$t_{n}=a+(n-1) d$
$\therefore \quad t_{19}=a+18 d=0$
$\therefore \quad a=-18 d$
$\therefore \quad \frac{t_{49}}{t_{29}}=\frac{a+48 d}{a+28 d}$
$=\frac{-18 d+48 d}{-18 d+28 d}=\frac{30 d}{10 d}=3$
$t_{49}: t_{29}=3: 1$