Question

the sum of 3 consecutive terms of an ap is 21 and the sum of the squares of these terms is 165.find these terms.

Answer 1

let the no be a-d,a,a+d
given that
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=18
d²=9
d=+3,-3

so the terms will be 4,7,10  or 10,7,4

Answer 2

Heya user,

Let the terms be ( a - d ), a, ( a + d )
Then, ATQ --> According to question --->

a - d + a + a + d = 21
=> 3a = 21 => a = 7;

Now, (a-d)² + a² + (a+d) = 3a² + 2d² = 165;
=> 3 * 7²  + 2d² = 165
=> 49 * 3 + 2d²  = 165
=> 147 + 2d² = 165
=> 2d² = 18
=> d = +3 or -3

Hence, The no.s are ( 7 + 3 ), ( 7 ), ( 7 - 3 ) = 10, 7, 4 or vice versa.....

_________________☺__________________☺______________________