If 10 times the tenth term of an A.P is equal to 30 times the thirtieth term, then find its 40th term.

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Solution

The correct option is A
0

$10 \times t_{10} = 30 \times t_{30}$
$10 \times [a + (9 - 1) d] = 30 \times [a + (30 - 1) d]$
$10 \times (a + 8 d) = 30 \times (a + 29 d)$
$(10 a + 80 d) = (30 a + 870 d)$
$- 20 a = 790 d$
$- 10 a = 390 d$
$10 a + 390 d = 0$
$a + 39 d = 0$ = $40^{t h} t e r m$ = 0

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