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The sum of first 3 term is 27 and sum of their square is 293. Then find sum of first 15 terms.

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Answer 1

Step-by-step explanation:

Let the three numbers are (a-d), a, a+d.

The sum of three numbers in AP is 27.

(a-d)+a+(a+d)=27(a−d)+a+(a+d)=27

3a=273a=27

a=9a=9

The sum of those three numbers squares is 293.

(a-d)^2+a^2+(a+d)^2=293(a−d)

2+a^2 +(a+d)^2 =293

a^2-2ad+d^2+a^2+a^2+2ad+d^2=293

a^2 −2ad+d^2+a^2 +a^2 +2ad+d^2 =293

3a^2+2d^2=2933

a^2 +2d^2 =293

3(9)^2+2d^2=2933(9)

2 +2d^2 =293 (a=9)

243+2d^2=293

243 + 2d^2 =293

d^2=25d^2=25

d=5d=5

The common difference is 5. The AP is defined as

a-d,a,a+2d,a+3d,...a−d,a,a+2d,a+3d,...

4, 9, 14, 19,...4,9,14,19,...

Therefore the required AP is 4, 9, 14, 19....

Then, the sum of 15 terms is

4 + 9 + 14 + 19 + 24 + 29 + 34 + 39 + 44 + 49

= 285

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Answer 2

Answer:

Let three terms be a - d, a, a+d

a - d + a + a + d = 27

3a = 27

a = 27/3 = 9

(a - d)² + a² + (a + d)² = 293

81 - 18d + d² + 81 +18d + d² = 293

243 +2d² = 293

2d² = 293 - 243 = 50

d² = 50/2 = 25

d = 5

15th term = a + 14d = 9 + 14(5)

= 79

Sum of 15 terms [S15]

= (15/2)(9 + 79)

= (15/2)(88)

= 15(44)

= 660

Is it right???