83
In an A.P., $15^{th}$ term exceeds the $8^{th}$ term by $21$. If sum of first $10$ terms is $55$, then form the A.P.
Show SolutionHide Solution↓
Sol. Let first term = $a$ and common difference = $d$
$(a + 14d) = (a + 7d) + 21$ (1 Mark)
$\Rightarrow d = 3$ (1/2 Mark)
Also, $S_{10} = 55 = \frac{10}{2}[2a + 9 \times 3]$ (1/2 Mark)
$\Rightarrow a = -8$ (1/2 Mark)
$\therefore$ A. P. is $-8, -5, -2, ...$ (1/2 Mark)
$(a + 14d) = (a + 7d) + 21$ (1 Mark)
$\Rightarrow d = 3$ (1/2 Mark)
Also, $S_{10} = 55 = \frac{10}{2}[2a + 9 \times 3]$ (1/2 Mark)
$\Rightarrow a = -8$ (1/2 Mark)
$\therefore$ A. P. is $-8, -5, -2, ...$ (1/2 Mark)