Given x11,x21,………,xn1 are in A.P. and x1=4 and x21=20.
⇒a=41
Given, x21=20.
⇒41+20⋅d=201
⇒d=−1001
Given, xn>50.
⇒xn1<501
⇒41−100n−1<501⇒n>24
∵n=25
Now, i=1∑25(xi1)=225[2×41−1001×24]=413
JEE Main 2018 — Mathematics Algebra
Let x11,x21,…,xn1(xi=0 for i=1,2,….,n) be in A.P. such that x1=4 and x21=20. If n is the least positive integer for which xn>50, then i=1∑n(xi1) is equal to
Held on 16 Apr 2018 · Verified 6 Jul 2026.
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