Find the sum of terms of the AP: $34+32+30+\dots \dots +10$

The correct option is B. 286

The formula to find the $n^{th}$ term of an arithmetic progression is $t_n=a+(n-1)d$.

where,

'$a_1$' is the first term,

${}^{\prime }{d}^{\prime }$ is the common difference,

${}^{\prime }{n}^{\prime }$ is the no. of terms,

'$t_n$' is the $n^{th}$ term.

Given,

first term $a_1=34$

second term $a_2=32$

$n^{th}$ term $t_n=10$

common difference, $d=a_2-a_1=32-34=-2$

Substituting the given values in the equation to find the $n^{th}$ term, we get

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

$\Rightarrow (n-1)(-2)=-24$

$\Rightarrow n-1=\frac{24}{2}=12$

$\Rightarrow$$n=13$

Sum of n terms of an AP is given by,
$S_n=\frac{n}{2}$ (first term + last term)

$\Rightarrow S_{13}=\frac{13}{2}(34+10)$

$\Rightarrow S_{13}=\frac{13}{2}\times 44$

$\Rightarrow S_{13}=13\times 22$

$\Rightarrow S_{13}=286$

Hence, the required sum is $286$.