Question

The sum of first 3 terms of an ap is 21 and sum of their squares is 155

Answer 1

Step-by-step explanation:

let the 3 terms a-d, a, a+d

sum of the terms is 21

a - d + a + a + d = 21

3a = 21

a = 21/3

a = 7

sum of their squares is 155

${(a - d)}^{2} + {a}^{2} + {(a + d)}^{2} = 155 \\ {a}^{2} - 2ad + {d}^{2} + {a}^{2} + {a}^{2} + 2ad + {d}^{2} = 155 \\ 3 {a}^{2} + 2 {d}^{2} = 155 \\ 3 {(7)}^{2} + 2 {d}^{2} = 155 \: \: \: \: (a = 7) \\ 3(49) + 2{d}^{2} = 155 \\ 147 + 2 {d}^{2} = 155 \\ 2{d}^{2} = 155 - 147 \\ 2 {d}^{2} = 8 \\ {d}^{2} = \frac{8}{2} = 4 \\ d = \sqrt{4} = 2$

a = 7 d = 2

the required AP is

a - d = 7 - 2 = 5

a = 7

a + d = 7 + 2 = 9

the first three terms of AP are 5, 7, 9

hope you get your answer