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

Correct answer: (C) -3. S=x+ka+x^2+k+2a+x^3+k+4a+x^4+k+6a+… S=x+x^2+x^3+x^4+...+ka+k+2a+k+4a+k+6a+… S=(xx^9-1)/(x-1)+a·9/22k+82 (xx^9-1)/(x-1)+a·9/22k+82=(x^10-x+9·ak+8x-1)/(x-1) So 45=9k+8⇒k=-3

Q41 / 75 · JEE Main 2020 (02 Sep Shift 2) · Mathematics

Let $S$ be the sum of the first $9$ term of the series :
$\{x + k a\} + \{x^{2} + (k + 2) a\} + \{x^{3} + (k + 4) a\} + \{x^{4} + (k + 6) a\} + \dots$ 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

A
$- 5$
B
$1$
Correct answer
C
$- 3$
D
$3$

Solution

$S = \{x + k a\} + \{x^{2} + (k + 2) a\} + \{x^{3} + (k + 4) a\} + \{x^{4} + (k + 6) a\} + \dots$

$S = (x + x^{2} + x^{3} + x^{4} + . . .) + (k a + (k + 2) a + (k + 4) a + (k + 6) a + \dots)$

$S = \frac{x (x^{9} - 1)}{(x - 1)} + a \cdot \frac{9}{2} (2 k + (8) (2))$

$\frac{x (x^{9} - 1)}{(x - 1)} + a \cdot \frac{9}{2} (2 k + (8) (2)) = \frac{x^{10} - x + 9 \cdot a (k + 8) (x - 1)}{(x - 1)}$

So $45 = 9 (k + 8) \Rightarrow k = - 3$

Hence, the correct answer is option: 'C'.

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Let S be the sum of the first 9 term of the series :x+ka+x2+k+2a+x3+k+4a+x4+k+6a+… where a≠0 and x ≠ 1. If S=x10-x+45ax-1x-1 , then k is equal to

Solution

Relevant Questions

Let $S$ be the sum of the first $9$ term of the series :
$\{x + k a\} + \{x^{2} + (k + 2) a\} + \{x^{3} + (k + 4) a\} + \{x^{4} + (k + 6) a\} + \dots$ 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