Question

Insert $6$ numbers between $3$ and $24$ such that the resulting sequence is an $A.P.$

Answer 1

Step 1 : Finding commom difference.
Let $A_1,A_2,A_3,A_4,A_5$ and $A_6$ be six numbers between $3$ and $24$ such that
$3,A_1,A_2,A_3,A_4,A_5,A_6,24$ are in A.P.
Here, $a=3,b=24,n=8$ and $d=$ common difference

As we know,
$d=\frac{b-a}{n-1}=\frac{24-3}{8-1}=\frac{21}{7}=3$
$\Rightarrow d=3$

Step $2:$ Finding numbers between $3$ and $24$.
Thus, $A_1=a+d=3+3=6$
$A_2=a+2d=3+2\times 3=9$
$A_3=a+3d=3+3\times 3=12$
$A_4=a+4d=3+4\times 3=15$
$A_5=a+5d=3+5\times 3=18$
$A_6=a+6d=3+6\times 3=21$

Final answer: Hence, six numbers between $3$ and $24$ are $6,9,12,15,18,21$