If 3k- k- -k2-34- are the first three terms of a geometric progression- where k- and - is the greatest integer function- then the value of -r-110rk-2 is

3k, k, [k^2 34] are in G.P. ⇒ k^2=3k[k^2 34] ⇒ k=3[k^2 34] ⇒ k is an integer. ⇒ k^2 34 is also an integer. ⇒ 3k^2 k 102=0 ⇒ k=6 or k= 173 As k is an integer, k= ...

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Byju's Answer

Standard XI

Mathematics

Sigma n3

If 3k, k, [k2...

Question

If $3 k,$ $k,$ $[k^{2} - 34]$ are the first three terms of a geometric progression, where $k \in R^{+}$ and $[.]$ is the greatest integer function, then the value of $10 \sum r = 1 (r)^{k / 2}$ is

55

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385

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3025

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325

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Solution

The correct option is C $3025$
$3 k, k, [k^{2} - 34]$ are in G.P.
$\Rightarrow k^{2} = 3 k [k^{2} - 34]$
$\Rightarrow k = 3 [k^{2} - 34]$
$\Rightarrow k$ is an integer.
$\Rightarrow k^{2} - 34$ is also an integer.
$\Rightarrow 3 k^{2} - k - 102 = 0$
$\Rightarrow k = 6$ or $k = - \frac{17}{3}$
As $k$ is an integer, $k = 6$

Now, $10 \sum r = 1 (r)^{k / 2} = 10 \sum r = 1 (r)^{3}$
$= 1^{3} + 2^{3} + \dots + 10^{3}$
$= {(\frac{10 \times 11}{2})}^{2} = 3025$

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Similar questions

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If 3k, k, [k2−34] are the first three terms of a geometric progression, where k∈R+ and [.] is the greatest integer function, then the value of 10∑r=1(r)k/2 is

The correct option is C 30253k, k, [k2−34] are in G.P. ⇒k2=3k[k2−34] ⇒k=3[k2−34] ⇒k is an integer. ⇒k2−34 is also an integer. ⇒3k2−k−102=0 ⇒k=6 or k=−173 As k is an integer, k=6 Now, 10∑r=1(r)k/2=10∑r=1(r)3 =13+23+⋯+103 =(10×112)2=3025

If $3 k,$ $k,$ $[k^{2} - 34]$ are the first three terms of a geometric progression, where $k \in R^{+}$ and $[.]$ is the greatest integer function, then the value of $10 \sum r = 1 (r)^{k / 2}$ is