Three numbers are inarithemetic progression and their sum is 18 sum of their squares is 140 find the numbers
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let the 3 numbers be a-d, a, a+d
a-d+a+a+d=18
3a=18
a=6
[a-d]²+a²+[a+d]²=140
a²+d²-2ad+a²+a²+d²+2ad
=3a²+2d²=140
=3[6]²+2d²=140
=108+2d²=140
=2d²=32
=d²=32
=d=±4
so the three numbers are 2,6 and 10
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