Question

Sum of three consecutive terms of an Arithmetic Progression is 42 and their product is 2520.
Find the terms of the Arithmetic Progression

Answer 1

Step-by-step explanation:

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Answer 2

18,14,10 or 10,14,18

Step-by-step explanation:

Let three consecutive terms of an A.P

a,a+d,a+2d

According to question

$a+a+d+a+2d=42$

$3a+3d=42$

$3(a+d)=42$

$a+d=\frac{42}{3}=14$

$d=14-a$..(1)

$a\times (a+d)(a+2d)=2520$

$a\times 14(a+(2(14-a))=2520$

$a(a+28-2a)=\frac{2520}{14}=180$

$a(28-a)=180$

$28a-a^2=180$

$a^2-28a+180=0$

$a^2-18a-10a+180=0$

$a(a-18)-10(a-18)=0$

$(a-18)(a-10)=0$

$a-18=0$

$a=18$

$a-10=0$

a=10

When a=18

d=14-a=14-18=-4

When a=10

d=14-10=4

When a=18 and d=-4

a+d=18-4=14

a+2d=18+2(-4)=10

Then, three consecutive terms are

18,14,10

When a=10 and d=4

a+d=10+4=14

a+2d=10+2(4)=18

Hence, the consecutive terms of AP are

18,14,10 or 10,14,18

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https://brainly.in/question/11255431:Answered by Captain brainly