If the first four terms of an arithmetic sequence are $a, 2 a, b$ and $a - 6 - b$ for some numbers $" a "$ and $" b "$ , find the value of the $100^{t h}$ terms

- 100

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- 300

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150

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- 150

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Solution

The correct option is A $- 100$
$a, 2 a, b, (a - 6 - b), . . . .$ forms an arithmetic progression
Common difference $d = a_{2} - a_{1} = 2 a - a = a d = a_{3} - a_{2} = b - 2 a$
$a = b - 2 a 3 a = b (1) a_{4} = a + 3 d (a - 6 - b) = a + 3 a = 4 a a - 6 - b = 4 a f r o m (1) a - 6 - 3 a = 4 a - 6 = 6 a$
$a = - 1$ and $d = - 1$
$a_{100} = a + 99 d = a + 99 a = 100 a a_{100} = - 100$

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