We have to find the sum n terms of the AP 2,4,6,...
Here a=2,d=4−2=2.
Therefore, Required sum Sn=n/2 [2a+(n−1)d]=n/2 [2×2+(n−1)×2]
⇒S_n=n[2+n−1]
⇒S_n=n(n+1) ...(1)

Put n=20 in (1), we get
S₂₀=20(20+1)=420

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