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