The sum of first 7 terms of an A.P. is 10 and that of next 7 terms is 17. Find the progression.
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 $1, \frac{8}{7}, \frac{9}{7}, \frac{10}{7} \ldots$
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