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Question

The fourth term of an Arithmetic progression is 10 and the eleventh term of it exceeds three times the fourth term by 1. Find the sum of the first $20$ terms of the progression.

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Solution

Given fourth term of A.P is 10.⇒a4=10→(1), where, a4 is denoted by fourth term of an A.P .And also eleventh term exceeds three times the fourth term by 1.⇒a11=3a4+1→(2)then we should find S20=?We know, an=a1+(n−1)dUsing this formula in (1) and (2),a+3d=10→(3)a+10d=3(a+3d)+1⇒a+10d=3a+9d+1⇒10d−9d=3a−a+1⟹d=2a+1→(4)Putting (4) in (3),⇒a+3(2a+1)=10⇒a+6a+3=10⇒7a+3=10⇒7a=10−3⇒a=77∴a=1Substitute a=1 in equation(4), we getThus, d=3(1)+1=3+1∴d=4Now, Sn=n2(2a+(n−1)d)Substitute n=20,a=1,d=4 in above formula, we get S20=202(2×(1)+(20−1)×3)=202(2+(19×3))=10(2+57)=10(59)∴S20=590

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