Question

find four consecutive terms in an A.P. whose sum is -54 and sum of 1st and 3rd term is -30. Let four consecutive terms be a-d,a,a+d,a+2d​

Answer 1

Step-by-step explanation:

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find 4 terms and of an ap

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$let \: the \: missing \: terms \: be = \\ a \\ a - d \\ a + d \\ a + 2d$

given sum of these terms is -54.....

$(a - d) + (a ) + (a + d) + \\ (a + 2d) = - 54$

$a + 2d+ a + a + a = - 54$

$4a + 2d= - 54$

$sum \: of \: 1 \: and \: 3 \: term \: is \: - 30$

$(a - d) + (a + d) = - 30 \\ 2a = - 30 \\ a = - 15$

$value \: of \: d \: will \: be$

$4a + 2d = - 54 \\ - 60 + 2d = - 54 \\ 2d = 6 \\ d = 3$

so the terms are

-15-3,-15,-15+3,-15+6

which are

-18,-15,-13,-9