Question

the sum of three consecutive terms of an ap is 21 and the sum of the squares of these terms is 165. Find the terms.​

Answer 1

Answer:

Step-by-step explanation:

let three terms be a-d,a and a+d

sum=21

3a=21

a=7

sum of their squares=

165=a^2+(a+d)^2+(a-d)^2

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

165=3a^2+2d^2

165=147+2d^2

18=2d^2

9=d^2

d=3 or -3

substitute and get values

Answer 2

Question : The sum of three consecutive terms of an ap is 21 and the sum of the squares of these terms is 165. Find the terms

Step-by-step explanation:

Let 1st consecutive term =   a-d

Let 1st consecutive term =   a-dLet 2nd consecutive term =  a

Let 1st consecutive term =   a-dLet 2nd consecutive term =  aLet 3rd consecutive term = a+d

Here , you can conclude that d is the  common difference

so ,

According To problem:  

a-d + a+ a-d =21

3a= 21

a=7

Now let's square them all :

(a-d)²+a²+(a+d)²=165

a²-2ad+d²+a²+a²+2ad+d²=165

3a²+2d²=165

147+2d²=165

2d²=18

d²=9

d=3

Now put the values :

a - d = 7 - 3

a-d =  4

a + d = 7 + 3

a+d = 10

Hence , The terms are 4,7,10