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Find five consecutive terms in an A.P. whose
sum of second and fourth term is 28 and the
product of second term and fifth term is 216.
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Let the first five terms be a, a + d , a +2d , a + 3d , a + 4d
a + d + a + 3d = 28
2a + 4d = 28
==> a + 2d = 14
==> a = 14 – 2d ----------------- equation 1
According to the question
a2 * a5 = 216
Form 1
==> ( 14 – 2d + d ) ( 14 – 2d + 4d ) = 216
==> ( 14 – d ) ( 14 + 2d ) = 216
==> 196 + 28d – 14d – 2d^2 = 216
==> 196 + 14d – 2d^2 = 216
==> 2d^ 2 – 14d + 20
==> d^ 2 – 7d + 10
==> ( d – 2 ) ( d – 5 )
==> d = 2 and d = 5
put d = 2 in equation 1
a = 14 – 2 ( 2 )
a = 14 – 4
a = 10
when d = 2
A.P. = 10,12,14,16,18............ANSWER
put d = 5 in equation 1
a = 14 – 2 ( 5 )
a = 14 – 10
a = 4
when d = 5
A.P. 4,9,14,19,24................ANSWER
Both answers are possible
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