S=x+ka+x2+(k+2)a+x3+(k+4)a+x4+(k+6)a+…
S=(x+x2+x3+x4+...)+(ka+(k+2)a+(k+4)a+(k+6)a+…)
S=(x−1)x(x9−1)+a⋅29(2k+(8)(2))
(x−1)x(x9−1)+a⋅29(2k+(8)(2))=(x−1)x10−x+9⋅a(k+8)(x−1)
So 45=9(k+8)⇒k=−3
JEE Main 2020 — Mathematics Algebra
Let S be the sum of the first 9 term of the series :
x+ka+x2+(k+2)a+x3+(k+4)a+x4+(k+6)a+… where a=0 and x=1. If S=x−1x10−x+45a(x−1) , then k is equal to
Held on 2 Sept 2020 · Verified 6 Jul 2026.
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−3
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