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Solution

Find the number of common terms in:
Series 1: $17, 21, 25, \ldots, 817$
Series 2: $16, 21, 26, \ldots, 851$

Step 1: Identify both APs
Series 1: $a_1 = 17,\ d_1 = 4$
Series 2: $a_2 = 16,\ d_2 = 5$

Step 2: First common term
By observation, $21$ is the first common term (appears in both series).

Step 3: Find common difference of new AP
Common terms will themselves form an AP.

Common difference $= \text{LCM}(d_1, d_2) = \text{LCM}(4, 5) = 20$

So common terms form AP: $21, 41, 61, \ldots$
with $a = 21,\ d = 20$

Step 4: Find the last common term
Last term must be $\leq$ smaller of the two last terms:
$\min(817, 851) = 817$

$T_n = a + (n-1)d \leq 817$
$21 + (n-1) \times 20 \leq 817$
$(n-1) \times 20 \leq 796$
$n - 1 \leq 39.8$
$n \leq 40.8$

So $n = 40$