Question

Sum of three consecutive terms of an Arithmetic Progression is 42 and their product is 2520.
Find the terms of the Arithmetic Progression.
(See Lesson 7)​

Answer 1

Answer:

Hey mate.

Let the 3 consecutive numbers in A. P be a+d,a,a-d.

case1

a-d+a+a+d=42

3a=42

a=42/3

a=14

case2

a*a-d*a+d=2520

14(14-d) (14+d)=2520

(14+d)(14-d) =2520/14

=180

14*14-d*d=180

196-d*d=180

-d*d=-16

(taking roots on bth sides)

d=4

the terms are 14-4,14,14+4

=10,14,18....

Hope this helps.....

Answer 2

Step-by-step explanation:

Let sum of three consecutive terms of an arithmetic progression will be a-b, a, a+b.

from given condition

(a-b) +a+(a+b)=42

a-b+a+a+b=42

3a=42

a=42/3

a=14

and

(a-b) *a*(a+b)=2520

(14-b) *14*(14+b)=2520

(14-b) (14+b)=2520/14

196-b²=180

-b²=180-196

-b²=-16

b²=16

b=4

terms are

a-b=14-4=10

a=14

a+b=14+4=18

the terms of the arithmetic progression is 10,14,18