Question

the sum of three terms of an arithmetic progression is 21 their products of first and the third terms exceed the second term by 6.find the numbers

Answer 1

the number is 1,7,13 this is arithmetic series a=1,d=6

Answer 2

Let $a_{1}$ = a - d , $a_{2}$ = a and $a_{3}$  = a + d.
From question,
$a_{1}$ + $a_{2}$ + $a_{3}$ = 21
a - d + a + a + d = 21
3a = 21
a = 7
Now, 
$a_{1}$ x $a_{3}$  = $a_{2}$ + 6
(a - d)(a + d) = a + 6
(7 - d) (7 + d) = 7 + 6
$7^{2}$ - $d^{2}$ = 13
49 - $d^{2}$ = 13
- $d^{2}$ = 13 - 49
- $d^{2}$ = - 36
$d^{2}$ = 36
d = $\sqrt{36}$ 
d = 6.
Now,
$a_{1}$ = a - d = 7 - 6 = 1
$a_{2}$ = a = 7
$a_{3}$ =  a + d. = 7 + 6 = 13
∴  $a_{1}$ = 1
    $a_{1}$ = 7
    $a_{1}$ = 13
Please comment if answer is satisfying/