sum of 3 consevutive terms is 21 in A.P. if the sum of square of these terms is 165 find these terms
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no be a,a+d,a+2d ,given a+a+d+a+2d=21 ,3a+3d =21,a+d =7 ,a2+a2+d2+2ad+a2+4d2+4ad = 165 ,3a2+5d2 +6ad = 165 solve these equation and find values
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Hey buddy! Here's your answer!
Let the first term be a, and common difference d.
Let the terms be a - d, a, a + d
a - d + a + a + d = 21
3a = 21
a = 7
(a - d)² + a² + (a + d)² = 165
(7 - d)² + 7² + (7 + d)² = 165
49 - 14d + d² + 49 + 49 + 14d + d² = 165
147 + 2d² = 165
2d² = 18
d² = 9
d = +- 3
Terms are 4, 7, 10 or 10, 7, 4
Hope this helps! :)
AccioNerd:
Pl mark as brainliest!
Let us take the numbers as a+d,a,a-d are in ap
It's the same thing, coz algebra always gives you the right answer
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