Question
How many term are there in the AP $41, 38, 35,....8?$
✓
Answer
In the given AP, we have $a = 41$ and $d = 38 - 41 = -3$
Suppose there are n term in the given AP, then
$T_n = 8 \Rightarrow a + (n - 1)d = 8$
$\Rightarrow 41 + (n - 1)(-3) = 8$
$\Rightarrow 41 - 3n + 3 = 8$
$\Rightarrow -3n = -36 \Rightarrow n = 12$
Hence there are $12$ terms in the given AP
Suppose there are n term in the given AP, then
$T_n = 8 \Rightarrow a + (n - 1)d = 8$
$\Rightarrow 41 + (n - 1)(-3) = 8$
$\Rightarrow 41 - 3n + 3 = 8$
$\Rightarrow -3n = -36 \Rightarrow n = 12$
Hence there are $12$ terms in the given AP
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