Question

The sum of first 3 terms of an A.P. is 27 and the
sum of their squares is 293. Find common
difference of A.P.​

Answer 1

Answer:

Step-by-step explanation:

Let the 3 terms be " a,a-d,a-2d"

According to the sum

The sum of first 3 terms =a+(a+d)+(a+2d)=27

a+a+d+a+2d=27

3a+3d=27

3 (a+d)=27

a+d=27/3

a+d=9

a=9-d (This is the first equation )

Again,according to the sum

The sum of their squares

a square +(a+d)whole square +(a+2d)whole square =293

a square +(a square +d square +2×a×d)+[a square +(2d)whole square +2×a×2d]=293

a square +(a square +d square +2ad)+(a square +4d square +4ad)=293

a square +a square +d square +2ad+a square +4d square +4ad=293

3a square +5d square +6ad=293 (This is the second equation )

Now, substituting "a" value in second equation,we get

3 (9-d)whole square +5d square +6 (9-d)d=293

3 (9 square -d square +2×9×d)+5d square +6 (9d-d square )=293

3 (9 square -d square +18d)+5d square +54d-6d square =293

3 (81+d square -18d)+5d square +54d-6d square =293

243+3d square -54d+5d square +54d-6d square =293

2d square =293-243

2d square =50

d square =50/2

d square =25

d=square root of 25

d=5

Therefore the common difference of A.P. =d=5 is the answer.

Answer 2

Let the first three numbers be

(a-d),(a^2),(a+d)

• (a-d)+(a^2)+(a+d)=27
• 3a=27
• a=27/3
• a=9

The sum of the square of three numbers is 293

• (a-d)^2+(a)^2+(a+d)^2=293
• a^2-2ad+d^2+a^2+a^2+2ad+d^2=293
• 3a^2+2d^2=293 (a=9)
• 3(9)^2+2d^2=293
• 243+2d^2=293
• 2d^2=293-243
• 2d^2=50
• d^2=50/2
• d^2=25
• d=5

The common difference of A.P is 5

The A.P is defined as

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

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

Therefore the required A.P is 4,9,14,19