Question
Find the sums given below : 34 + 32 + 30 + … + 10
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Answer
$\begin{aligned} & 34+32+30+\ldots+10 \\ & \text { Here, } a=34, d=32-34=-2, I=10 \\ & T_n=a+(n-1) d \\ & 10=34+(n-1)(-2) \\ & -24=-2(n-1) \\ & =\frac{-24}{-2} \\ & =n-1 \\ & =12 \\ & \therefore n=12+1=13 \\ & S_n=\frac{n}{2}[a+l] \\ & =\frac{13}{2}[34+10] \\ & =\frac{13}{2} \times 44 \\ & =286 .\end{aligned}$
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