Let S be the sum of the first 9 terms of the series- x-ka - x2-k-2a - x3-k-4a- x4-k-6a - where a- 0 and a- 1- If S-x10-x-45ax-1x-1- then k is equal to-

S={ x+ka }+{x^2+(k+2)a }+{ x^3+(k+4)a } up to 9 terms S=(x+x^2+...+x^9 )+a{ k+(k+2)+(k+4)+....... up to 9 terms) } S=x(1 x^9 )1 x+a{ 9k+(2)(36) } S=x^10 xx 1+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Distance Formula Using Complex Numbers

Let S be the ...

Question

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

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $- 3$
$S = {x + k a} + {x^{2} + (k + 2) a} + {x^{3} + (k + 4) a}$ up to $9$ terms
$S = (x + x^{2} + . . . + x^{9}) + a {k + (k + 2) + (k + 4) + . . . . . . . up to 9 terms)}$
$S = \frac{x (1 - x^{9})}{1 - x} + a {9 k + (2) (36)}$
$S = \frac{x^{10} - x}{x - 1} + 9 a k + 72 a$
$S = \frac{x^{10} - x + 45 a (x - 1)}{x - 1} = \frac{x^{10} - x + (9 k + 72) a (x - 1)}{x - 1}$
$\Rightarrow 45 = 9 k + 72$
$\Rightarrow 9 k = - 27$
$\Rightarrow k = - 3$

Suggest Corrections

Similar questions

Q. Let

S_{k}

be the sum of an infinite GP series whose first term is

k

and common ratio is

\frac{k}{k + 1} (k > 0)

. Then, the value of

\infty \sum k = 1 \frac{(- 1)^{k}}{S_{k}}

is equal to

Q. If

S_{1} = \sum_{k = 0}^{n} {^{n} C}_{k} cos (k x) cos (n - k) x

and

S_{2} = \sum_{k = 0}^{n} {^{n} C}_{k} sin (k x) sin (n - k) x

, then

S_{1} - S_{2}

equals

Q. If

\int \frac{x^{2} - 1}{x^{4} + x^{2} + 1} d x = \frac{1}{2} ln ∣ ∣ ∣ \frac{x^{2} - x + K}{x^{2} + x + K} ∣ ∣ ∣ + C

, Then value of

K

is (where

K

is fixed constant and

C

is integration constant)

Q. Let x > 0, y > 0, z > 0 are respectively the

2^{n d}, 3^{r d}, 4^{t h},

terms of a G.P and

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{k} & x^{k + 1} & x^{k + 2} y^{k} & y^{k + 1} & y^{k + 2} z^{k} & z^{k + 1} & z^{k + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (r - 1)^{2} (1 - \frac{1}{r^{2}})

(where r is the common ratio) then

Q. Sum of the series ,

7 + 77 + 777 + . . . . .

n

terms be

S = (\frac{7}{m}) (k^{n + 1} - (k - 1) n - k)

.Find

\frac{m - k - 1}{k}

Join BYJU'S Learning Program

Let S be the sum of the first 9 terms of the series: {x+ka}+{x2+(k+2)a}+{x3+(k+4)a}+{x4+(k+6)a}+... where a≠0 and a≠1. If S=x10−x+45a(x−1)x−1, then k is equal to:

The correct option is B −3S={x+ka}+{x2+(k+2)a}+{x3+(k+4)a} up to 9 terms S=(x+x2+...+x9)+a{k+(k+2)+(k+4)+.......up to 9 terms)} S=x(1−x9)1−x+a{9k+(2)(36)} S=x10−xx−1+9ak+72a S=x10−x+45a(x−1)x−1=x10−x+(9k+72)a(x−1)x−1 ⇒45=9k+72 ⇒9k=−27 ⇒k=−3

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