Find the sums given below:
$34 + 32 + 30 + … + 10$

 Given:

$34 + 32 + 30 + … + 10$

To do:

We have to find the given sum.

Solution:

$a=34, d=32-34=-2, a_n=10$

We know that,

$a_n=l=a+(n-1)d$

$S_{n}=\frac{n}{2}[a+l]$

$a_{n}=a+(n-1) d$

$10=34+(n-1)(-2)$

$10-34=(n-1)(-2)$

$\frac{-24}{-2}=n-1$

$n-1=12$

$n=12+1$

$n=13$

Therefore,

$S_{13}=\frac{13}{2}[34+10]$

$=\frac{13}{2}(44)$

$=13 \times 22$

$=286$

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Updated on: 10-Oct-2022

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