Question

if the sum of 4 numbers in AP is 24,and the sum of their squares is 164, find no's​

Answer 1

Answer:

$the \: required \: numbers \: are \: \: \: \: \: \: 3 \: \: \: \: 5 \: \: \: \: \: 7 \: \: \: \: \: 9$

Step-by-step explanation:

$a - 3d \: \: \: \: a - d \: \: \: \: \: \: a + d \: \: \: \: a + 3d \\ \\ 4a = 24 \\ \\ a = 6 \\ \\ (6 - 3d) \: \: \: \: \: 6 - d \: \: \: \: \: \: 6 + d \: \: \: \: \: 6 + 3d \\ \\ \\ \\ \\ {(6 - 3d)}^{2} + \: \: \: \: \: \: {(6 - d)}^{2} \: \: + \: \: \: \: {(6 + d)}^{2} \: \: + \: \: \: {(6 + 3d)}^{2} = 164 \\ \\ 4( {6)}^{2} + 9 {d}^{2} + 2 {d}^{2} + 9 {d}^{2} - 36d - 12d + 12d + 36d = 164 \\ \\ 144 + 20 {d}^{2} = 164 \\ \\ 20 {d}^{2} = 20 \\ \\ d = 1 \\ \\ the \: required \: numbers \: are \: \: \: \: \: \: 3 \: \: \: \: 5 \: \: \: \: \: 7 \: \: \: \: \: 9$