The sum of first 7 terms of an A.P. is 10 and that of next 7 terms is 17.

Solution:

We have:

$S_{7}=10$

$\Rightarrow \frac{7}{2}[2 a+(7-1) d]=10$

$\Rightarrow \frac{7}{2}[2 a+6 d]=10$

$\Rightarrow a+3 d=\frac{10}{7} \ldots(\mathrm{i})$

Also, the sum of the next seven terms $=S_{14}-S_{7}=17$

$\Rightarrow \frac{14}{2}[2 a+(14-1) d]-\frac{7}{2}[2 a+(7-1) d]=17$

$\Rightarrow 7[2 a+13 d]$

$-\frac{7}{2}[2 a+6 d]=17$

$\Rightarrow 14 a+91 d-7 a-21 d=17$

$\Rightarrow 7 a+70 d=17$

$\Rightarrow a+10 d=\frac{17}{7} \quad \ldots$ (ii)

From (i) and (ii), we get:

$\frac{10}{7}-3 d=\frac{17}{7}-10 d$

$\Rightarrow 7 d=1$

Putting the value in (i), we get:

$a+3 d=\frac{10}{7}$

$\Rightarrow a+\frac{3}{7}=\frac{10}{7}$

$\Rightarrow a=1$

$\therefore a=1, d=\frac{1}{7}$

The progression thus formed is $1, \frac{8}{7}, \frac{9}{7}, \frac{10}{7} \ldots$