Four consecutive terms in an arithmetic progression have the sum as $20$ while the sum of their squares is $120$ . Find the four consecutive terms.

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Solution

$a_{1} + a_{2} + a_{3} {+ a}_{4} = 20 {a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2} + {a_{4}}^{2} = 120 a + (a + d) + (a + 2 d) + (a + 3 d) = 20 a^{2} + {(a + d)}^{2} + {(a + 2 d)}^{2} + {(a + 3 d)}^{2} = 120 4 a + 6 d = 20 4 a^{2} + 14 d^{2} + 12 a d = 120 2 a + 3 d = 10 . . . . . . . . . . . . (1) 4 a^{2} + 14 d^{2} + 12 a d = 120 . . . . . . . . . . (2) s q u a r i n g (1) {(2 a + 3 d)}^{2} = 100 4 a^{2} + 9 d^{2} + 12 a d = 100.......... (3) (3) - (2) - 5 d^{2} = - 20 d = 2 a = 2 A P - > 2, 4, 6, 8, . . . .$

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