Explanation

(a)
Let common difference $d_{1} = - 3$ and first term be $a$.
$\therefore$ Series become
$a,a - 3,a - 6,\ldots,a - 27$
$\therefore S = 10a + ( - 3 - 6 - ... - 27)$
$\Rightarrow - 30 = 10a - 3(1 + 2 + ... + 9)$
$\Rightarrow - 30 = 10a - 3\left\lbrack \frac{9(9 + 1)}{2} \right\rbrack$
$\Rightarrow - 30 = 10a - 135$
$\Rightarrow 10a = 105\ \ \Rightarrow a = \frac{105}{10}$
Now, correct common difference $d_{2} = 3$
$\therefore$ Required sum = $\frac{10}{2}\left\lbrack 2 \times \frac{105}{10} + (10 - 1)3 \right\rbrack$
$= 5\left( \frac{105}{5} + 27 \right) = 5 \times 48 = 240$