Consider two G.P.’s. 2, 22, 23, …… and 4, 42, 43, …… of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is <math><mrow><msup><mrow><mrow><mo>(</mo><mrow><mn>2</mn></mrow><mo>)</mo></mrow></mrow><mrow><mfrac><mrow><mn>2</mn><mn>2</mn><mn>5</mn></mrow><mrow><mn>8</mn></mrow></mfrac></mrow></msup></mrow></math> , then <math><mrow><mstyle displaystyle="true"><munderover><mo>∑</mo><mrow><mi>k</mi><mo>=</mo><mn>1</mn></mrow><mrow><mi>n</mi></mrow></munderover><mrow><mi>k</mi><mrow><mo>(</mo><mrow><mi>n</mi><mo>−</mo><mi>k</mi></mrow><mo>)</mo></mrow></mrow></mstyle></mrow></math> is equal to:

Given G.P’s 2, 22, 23, …60 term and 4, 42, 43, … of 60Now G.M. = (2)2258⇒ (2, 22, 23, ....)160+n= (2)2258⇒n=578, 20 so n=20∴∑k=1nk (n−k)20×20×212−20×21×416=1330

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Maths Class 12th Sequence and Series Maths NCERT Exemplar Solutions Class 12th Chapter One

Consider two G.P.’s. 2, 2², 2³, …… and 4, 4², 4³, …… of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is ${(2)}^{\frac{225}{8}}$ , then $\sum_{k = 1}^{n} k (n - k)$ is equal to:

Option 1 -

560

Option 2 -

1540

Option 3 -

1330

Option 4 -

2600

2 Views | Posted 2 months ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Given G.P’s 2, 2², 2³, …60 term and 4, 4², 4³, … of 60

Now G.M. = ${(2)}^{\frac{225}{8}} \Rightarrow {(2, 2^{2}, 2^{3}, . . . .)}^{\frac{1}{60 + n}} = {(2)}^{\frac{225}{8}} \Rightarrow n = \frac{57}{8}, 20 s o n = 20$

$∴ \sum_{k = 1}^{n} k (n - k) 20 \times \frac{20 \times 21}{2} - \frac{20 \times 21 \times 41}{6} = 1330$

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Consider two G.P.’s. 2, 2², 2³, …… and 4, 4², 4³, …… of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is ${(2)}^{\frac{225}{8}}$ , then $\sum_{k = 1}^{n} k (n - k)$ is equal to:

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Consider two G.P.’s. 2, 22, 23, …… and 4, 42, 43, …… of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is (2)2258 , then ∑k=1nk(n−k) is equal to:

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Consider two G.P.’s. 2, 2², 2³, …… and 4, 4², 4³, …… of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is ${(2)}^{\frac{225}{8}}$ , then $\sum_{k = 1}^{n} k (n - k)$ is equal to: