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

Posted by Prudhvi Yadav 6 years ago

CBSE > Class 10 > Mathematics

2 answers

Gaurav Chaurasia 6 years ago

Let the three terms be (a-d),(a),and (a+d) Then,1st condition (a-d)+(a) +(a+d) = 21 adt at atel: 4 3a=21 a=7 2nd condition, (a-d)^2+(a)^2+ (a+d)^2 =165 After solving this equation we approach to d=+-3 So, 1st term =(a-d)=7-3=4 2nd term=a=7 3rd term =(a+d)=7+3=10 -By Srinivasa Gaurav Chaurasia

2Thank You

Khshsj Dhbdbdbdbjs 6 years ago

Let the 3 numbers be a-d,a,a+d Sum=3a=21 So,a=7 Since,the sum of squares is 165 Therefore,(a-d)²+(a)²+(a+d)² =49+d²-14d+49+49+d²+14d=165 =2d²+147=165 =2d²=165-147=18 d²=9 d=3 Therefore,the given A.P is 7-3,7,7+3 Which is 4,7,10

3Thank You

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