Question
Find the sum of 34 + 32 + 30 + … + 10.
✓
Answer
Here, a = 34
d = 32 - 34 = -2
l = 10
Let the number of terms of the AP be n.
We know that,
l = a + (n - 1)d
$ \Rightarrow $ 10 = 34 + (n - 1) (-2)
$ \Rightarrow $ (n - 1) (-2) = - 24
$ \Rightarrow $ $n - 1 = \frac{{ - 24}}{{ - 2}} = 12$
$ \Rightarrow n = 13$
Again, we know that,
${S_n} = \frac{n}{2}(a + l)$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}(34 + 10)$
$ \Rightarrow {S_{13}} = 286$
Hence, the required sum is 286.
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