Let S be the sum of the first 9 terms of the series: \x+ ka\+\x^2+ (k+2 )a\+\x^3+ (k+4 )a\+\x^4+ (k+6 )a\+ … where a 0 and x 1 If S= x^10-x+45a (x-1 )x-1 then k is equal to :

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Correct answer: A. (a)JEE Main 2020 Seires (x+ ka )+ (x^2+ (k+2 )a )+ … …9 terms) ⇒S= (x+x^2+x^3+ … …9 terms +a [k+ (k+2 ) + (k+4 )+ … .9terms ] ⇒S= x (x^9-1 )x-1+ 92 [2 ak+8 (+2a ) ] ⇒S= x^10-xx-1+ 9 ka+72a1= x^10+45a (x-1 )x-1 (given) ⇒ x^10-x+9a (k+8 ) (x-1 )x-1= x^10-x+45a (x-1 )x-1 ⇒9a (k+8 )=45a ⇒k+8=5 ⇒k=-3

$Let S be the sum of the first 9 terms of the series: {x + ka} + {x^{2} + (k + 2) a} + {x^{3} + (k + 4) a} + {x^{4} + (k + 6) a} + \dots where a \neq = 0 and x \neq = 1 If S = \frac{x ^{10} - x + 45 a ( x - 1 )}{x - 1} then k is equal to :$

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