Question

Find an ap whose sum of first three terms is 21 and the sum of their squares is 155

Answer 1

hi dear,
here is ur answer..............................
let the terms of A.P
  1st term=a
   2nd term=a+d
  3rd term=a+2d
according to the question......
a+a+d+a+2d=21
3a+3d=21
a+d=7
d=7-a........................................................................................1.
a^2+(a+d)^2+(a+2d)^2=155
a^2+(a+7-a)^2+(a+14-2a)^2=155
a^2+49+196-28a+a^2=155
2a^2-28a+245=155
2a^2-28a+90=0
a^2-14a+45=0
a^2-9a-5a+45=0
a(a-9)-5(a-9)=0
(a-5)(a-9)=0
so,
if a=5 then d=2
if a=9 then d=-2
so A.P formed :
5,7,9....
or
9,7,5......
HOPE IT HELPS UH...........................
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Answer 2

Answer:

let the terms of A.P

 1st term=a

  2nd term=a+d

 3rd term=a+2d

according to the question......

a+a+d+a+2d=21

3a+3d=21

a+d=7

d=7-a........................................................................................1.

a^2+(a+d)^2+(a+2d)^2=155

a^2+(a+7-a)^2+(a+14-2a)^2=155

a^2+49+196-28a+a^2=155

2a^2-28a+245=155

2a^2-28a+90=0

a^2-14a+45=0

a^2-9a-5a+45=0

a(a-9)-5(a-9)=0

(a-5)(a-9)=0

so,

if a=5 then d=2

if a=9 then d=-2

so A.P formed :

5,7,9....

or

9,7,5......

Step-by-step explanation: