Find the sums given below:
$ 7+10 \frac{1}{2}+14+\ldots+84 $
Given:
\( 7+10 \frac{1}{2}+14+\ldots+84 \)
To do:
We have to find the given sum.
Solution:
In the given series,
First term $a=7$
Common difference $d=10\frac{1}{2}-7=\frac{10\times2+1-7\times2}{2}=\frac{21-14}{2}=\frac{7}{2}$
Last term $l=a+(n-1)d=84$
$84=7+(n-1)\times\frac{7}{2}$
$84-7=(n-1) \times\frac{7}{2}$
$2\times77=7n-7$
$7n=154+7$
$n=\frac{161}{7}$
$n=23$
Sum of the given series is $\frac{n}{2}(a+l)$.
$S_n=\frac{23}{2}\times(7+84)$
$=\frac{23}{2}\times91$
$=\frac{2093}{2}$
$=1046\frac{1}{2}$
Therefore, $7+10 \frac{1}{2}+14+\ldots+84=1046\frac{1}{2}$.
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