Question

The sum of first three terms of an arithmetic
progression is 24 and the sum of their squares is 224.find the first three terms of this arithmetic progression.

Answer 1

Answer:4,8,12 or 12,8,4

Step-by-step explanation:

Let a-d = frst term

a=second term

a+d=third term

A+A+d+A-d=24

3a=24

A=8

(A+d)^2 +a^2 + (a-d)^2 =224

64+d^2+16d+64+64+d^2-16d=224

2(96+d^2)=224

96+d^2=112

D^2=16

D=+4 , -4

AP..... 12,8,4 [A=8 & D=-4]

AP.......4,8,12 [A=8 & D=4]

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Answer 2

let the three terms of ap be a-d, a,. a+d

a-d+a+a+d=24

3a=24

a=24/3

a=8

(a-d)2+(a)2+(a+d)2=224

a2+d2-2ad+a2+a2+d2+2ad=224

3a2+2d2=224

putting a value

102+2d2=224

2d2=224-192

2d2=32

d2=16

d=4

then AP will be 8,12,16....... ans

hope it helful