Question:
Find the sum of 16 terms of the AP $6,5 \frac{1}{3}, 4 \frac{2}{3}, 4, \ldots$
Solution:
To find: Sum of 16 terms of the AP
Given:
First term = 6
Common difference $=-\frac{2}{3}$
$\Rightarrow S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{n}=\frac{16}{2}\left[2 \times 6+15 \times\left(-\frac{2}{3}\right)\right]$
$\Rightarrow S_{n}=\frac{16}{2}[12-10]_{S_{n}}=16$
The sum of first 16 terms of the series is 16
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