Question

find four consecutive term in AP whose sum is 20 and the sum of whose square is 120​

Answer 1

let the four terms be a–d, a, a+d and a+2d

now according to question

a–d + a + a+d + a+2d = 20

4a+2d = 20

2a + d = 10

that is d = 10–2a

now it also says that

(a–d) ka square + a Ka square + (a+d) Ka square + (a+2d) Ka square = 120

solving this We get 2a×a + 3d×d + 2×a×d = 60 [put d=10–2a]

then you find a = ......

similarly d = .......

and then all the terms

Answer 2

Here Is Your Answer My Friend ✌️☺️

a = 5.

d = 1.

Hope It Help You..

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@ Nishu...❣️❣️❣️