Question

In an A. P the sum of four terms is 20 and the sum of their squares is 120.Find the four terms​

Answer 1

Step-by-step explanation:

$\sf \: Let \: the \: four \: number \: in \: A.P \: be \\ \tt \: a - 3d , \: a - d , \: a + b , \: a + 3d \: \: \: \: \: - - - (1)$

$\sf \: Given \: that \: sum \: of \: the \: terms = 20$

$\tt \implies \: (a - 3d) + (a - d)(a + d) + (a + 3d) = 20$

$\bf \: \: 4a \: = 20 \\ \bf a = 5 \: \: \: \: \: - - - (2)$

$\sf \: Given \: that \: sum \: of \: square \: of \: the \: terms = 120$

$\tt \implies \: (a - {3d)}^{2} + (a - d {)}^{2} + {(a + d) }^{2} + (a + 3d {)}^{2} = 120$

$\tt \implies \: ( {a}^{2} + {9d}^{2} - 6ad) + ( {a}^{2} + {d}^{2} - 2ab) + ( {a}^{2} + {d}^{2} + 2ad) + ( {a}^{2} + {9d}^{2} + 9ad) = 120$

$\tt \implies \: {4a}^{2} + {20d}^{2} = 120$

$\bf \: Substitute \: \: \green{a = 5} \: from \: \: \: \: \: \: \: \: \: - - (2)$

$\tt \implies \: 4 {(5)}^{2} + {20d}^{2} = 120$

$\tt \implies100 + {20d}^{2} = 120$

$\tt \implies {20d}^{2} = 120 - 100$

$\tt \implies {20d}^{2} = 20$

$\blue{ \boxed{\bf \implies \: d \: = + 1 \: \: \: or \: \: \: \: - 1}}$

$\bf\red{Since \: A.P \: cannot \: be \: negative }$

$\bf \green{Substitute \: } \: a = 5 \green{ \: \: and} \: \: d = 1 \green{ \: \: in \: (1), \: we \: get}$

$\sf \red{a - 3d}, \: \green{a - d}, \: \pink{a + b}, \: \blue{a + 3d} = 2,4,6,8$