Question

the sum of three consecutive term of an AP is 21 and the sum of the squares of these terms is 155.find the terms.​

Answer 1

Step-by-step explanation:





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Question



The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165: Find these terms.

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Solution



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Let the numbers be:

a−d,a,a+d

a−d+a+a+d=21

3a=21

∴a=7

(a−d)2+a2+(a+d)2=165

a2+d2−2ad+a2+a2+d2+2ad=165

3a2+2d2=165

2d2=165−3(7)2=18

d2=9

∴d=3

∴ the numbers are

a−d,a,a+d=7−3,7,7+3=4,7,10

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Answer 2

Answer:

Let the required terms be (a-d), a and (a+d). then (a-d)+a+(a+d)=3a=21 a=7

also (a-d) whole square +a square + (a+d) whole square =165

3a square + 2d square =165

(3×49+2d square) = 165

2d square= 165-147= 18

d square = 9

d=+ (or)-3

thus a= 7 d= +(or) -3

hence, the required terms are (4, 7 ,10) or (10,7,4)