In a given A.P. $a = 8, T_{n} = 33, S_{n} = 123$ . Find $d$ and $n$ .

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Solution

Assuming $T_{n}$ as the last term $l$
For the given A.P $a = T_{1} = 8, l = T_{n} = 33$ and $S_{n} = 123$
Now, $S_{n} = \frac{1}{2} n [a + l]$
$123 = \frac{n}{2} (8 + 33)$
$123 = \frac{41}{2} n$
$123 \times \frac{2}{41} = n$
$⟹ n = 6$

Now,
$T_{n} = a + (n - 1) d$
$T_{6} = a + 5 d$
$33 = 8 + 5 d$
$25 = 5 d$
$d = 5$

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