Solve these three equations to get the value of $a$ , $n$ , $d$ .
$(2 n - 1) d = 10.5 . . . . (i)$
$n (a + (n - 1) d) = 2 d . . . . . . (i i)$
$n (a + n d) = 30 . . . . . . . (i i i)$

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Solution

Given,
$(2 n - 1) = 10.5......... (i)$
$n (a + (n - 1) d) = 2 d . . . . . . . . . (i i)$
$n (a + n d) = 30............ (i i i)$

$e q (i i i) - e q (i i)$ gives
$n d = 30 - 2 d$
$\Rightarrow n d + 2 d = 30........ (i v)$

$2 \times e q (1),$ gives $4 n d - 2 d = 21......... (v)$

$e q (i v) + (v)$ , gives $5 n d = 51$

$2 \times e q (i v) . . . . e q (i)$ , gives $5 d = 49.5 \Rightarrow d = 9.9$

$5 n d = 51$

$\Rightarrow n = \frac{51}{5 d} = \frac{51}{5 \times 9.9} = \frac{51}{49.5} = 1.03$

$n = 1.03$

putting the values of $n & d$ in eq (iii), we get

$a = \frac{30}{n} - n d$

$= \frac{30}{1.03} - \frac{51}{5}$

$= 29.12 - 10.2 = 18.92$

$a = 18.92, n = 1.03, d = 9.9$

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