Question

in AP the sum I of three terms is 60.sum of three squares is 1250.find middle term?
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Answer 1

Answer:

middle term = 20

Step-by-step explanation:

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

a-d is the first term

a is middle term

a+d is the third term

sum of the three terms = 60

a - d + a + a + d = 60

3a = 60

a = 60/3

a = 20

sum of squares of three terms = 1250

${(a - d)}^{2} + {a}^{2} + {(a + d)}^{2} = 1250 \\ {a}^{2} - 2ad + {d}^{2} + {a}^{2} + {a}^{2} + 2ad + {d}^{2} = 1250 \\ 3 {a}^{2} + 2 {d}^{2} = 1250 \\ 3 {(20)}^{2} + 2 {d}^{2} = 1250 \\ 3(400) + 2 {d}^{2} = 1250 \\ 1200 + 2 {d}^{2} = 1250 \\ 2 {d}^{2} = 1250 - 1200 \\ 2 {d}^{2} = 50 \\ {d}^{2} = \frac{50}{2} \\ {d}^{2} = 25 \\ d = \sqrt{25} \\ d = 5$

a = 20

d = 5

the AP is

a - d = 20 - 5 = 15

a = 20

a + d = 20 + 5 = 25

the AP is 15, 20, 25

Hope you get your answer