Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Let $A_{1}, A_{2}, A_{3}, A_{4}$, and $A_{5}$ be five numbers between 8 and 26 such that

$8, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 26$ is an A.P.

Here, $a=8, b=26, n=7$

Therefore, $26=8+(7-1) d$

$\Rightarrow 6 d=26-8=18$

$\Rightarrow 6 d=26-8=18$

$\Rightarrow d=3$

$A_{1}=a+d=8+3=11$

$A_{2}=a+2 d=8+2 \times 3=8+6=14$

$A_{3}=a+3 d=8+3 \times 3=8+9=17$

$A_{4}=a+4 d=8+4 \times 3=8+12=20$

$\mathrm{A}_{5}=a+5 d=8+5 \times 3=8+15=23$

Thus, the required five numbers between 8 and 26 are $11,14,17,20$, and 23 .