Question

find 5 consecutive terms in an A.P. whose sum of 2nd and 4th term is 28 and product of 2nd and 5th term is 216
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Answer 1

Let the first 5 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 question

a2 * a5 = 216

=> a+d * a + 4d = 216

=>From 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 eq 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 eq 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

HOPE IT HELPED U

Answer 2

Answer:

$ANSWER</p><p></p><p>Let the terms of AP be</p><p></p><p>A−2d,a−d,a,a+d,a+2d</p><p></p><p>A/Q</p><p></p><p>(a−2d)+(a−d)+a+a(a+d)+(a+2d)=605a=60⇒a=12</p><p></p><p>and a×(a+d)=172+a+2d⇒a2+ad=172+a+2d⇒144+12d=172+12+2d⇒12d−2d=184−144=40⇒10d=40⇒d=4∴a=12,d=4</p><p></p><p>and the terms are</p><p></p><p>12−4×2,12−4,12,12+4,12+2×4=4,8,12,16,20</p><p></p><p>$

this is your answer dear