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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 $S={x^{10}-x+45a(x-1)10\over x-1}$, then k is equal to

(1) –3 

(2) 1 

(3) –5 

(4) 3

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Ans. (1) –3

Sol. ${x(x^9-1)\over(x-1)}+a.{9\over2}[2k+8.2]$

$={x^{10}-x+9.a(k+8)(x-1)\over(x-1)}$

so 45 = 9(k + 8) $\implies$ k = –3

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