Question

four numbers are in AP the sum of the first three is 21 and the sum of last three is 30 find the numbers

Answer 1

let us consider ap terms as
 
a-d,a,a+d,a+2d
sum of first three no
 a−d+a+a+d=21⟹3a=21⟹a=7
sum of last three is 30
3a+3d=30
21+3d=30
d=3
therfore terms are

4,7,10,13

Answer 2

Here is your solution

Given :-

Sum of first 3 numbers = 21

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

⇒ 3a + 3d = 21

⇒a + d = 7 ---------------( 1 )

The sum of last three number is 30

⇒ a + d + a + 2d + a + 3d = 30

⇒3a + 6d = 30

⇒a + 2d = 10--------------( 2 )

on subtracting equation ( 1 ) from ( 2 )

a + d - a -2d = 7 -10

-d = -3

d = 3

putting the value of d in equation (1)

a + 3 = 7

a = 7 - 3

a = 4

d = 3

The numbers are
a = 4 , (a + d) = 7✔

(a + 2d) = 10✔

(a + 3d) = 13✔

AP are 4 , 7 , 10 , 13 .

Hope it helps you