Put five terms between the numbers $1$ and $1.3$ so that together with the given terms they will form an arithmetic progression (A.P).
Let the first term of A.P $= t_1 = a$
Let the common difference of A.P $= d$
Five terms are to be put between the numbers $1$ and $1.3$
$\implies$ The $1^{st}$ term of the required A.P $= a = 1$
and the $7^{th}$ term of the required A.P $= t_7 = 1.3$
Now, the $n^{th}$ term of A.P $= t_n = a + \left(n - 1\right) d$
$\therefore \;$ $7^{th}$ term of A.P $= t_7 = a + 6d = 1.3$ $\;\;\; \cdots \; (1)$
Substituting $a = 1$ in equation $(1)$ gives $\;\;$ $1 + 6d = 1.3$
i.e. $\;$ $6d = 0.3$ $\implies$ $d = 0.05$
$\therefore \;$ $2^{nd}$ term of A.P $= t_2 = a + d = 1 + 0.05 = 1.05$
$3^{rd}$ term of A.P $= t_3 = t_2 + d = 1.05 + 0.05 = 1.10$
$4^{th}$ term of A.P $= t_4 = t_3 + d = 1.10 + 0.05 = 1.15$
$5^{th}$ term of A.P $= t_5 = t_4 + d = 1.15 + 0.05 = 1.20$
$6^{th}$ term of A.P $= t_6 = t_5 + d = 1.20 + 0.05 = 1.25$
$\therefore \;$ The five terms to be put between the numbers $1$ and $1.3$ so that together with the given terms they will form an A.P are: $\;\;$ $1.05, \; 1.10, \; 1.15, \; 1.20 \;$ and $\; 1.25$