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Insert four number between 6 and 192 so ...

Insert four number between 6 and 192 so that the resulting sequence is a G.P.

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To solve the problem of inserting four numbers between 6 and 192 such that the resulting sequence forms a geometric progression (G.P.), we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Terms**: We need to insert four numbers between 6 and 192. Let's denote these four numbers as \( a, b, c, d \). The sequence will then be: \[ 6, a, b, c, d, 192 \] ...
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AAKASH INSTITUTE ENGLISH-SEQUENCES AND SERIES -Try Yourself
  1. If a, b, c are in G.P. , then show that (i) (a^(2) - b^(2))(b^(2)...

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  2. Find the G.M. between the numbers. (i) 6 and 216 " " (ii...

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  3. Find the G.P. between the numbers . 1 and (4)/(25) " " ...

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  4. Insert four number between 6 and 192 so that the resulting sequence is...

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  5. Insert six G.M. ' s between (8)/(27) and (-81)/(16) .

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  6. If one G.M., G and two A.M's p and q be inserted between two given num...

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  7. The sum of two numbers is 6 times their geometric means, show that nu...

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  8. If the A.M. between two positive numbers exceeds their G.M. by 2 and t...

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  9. If G(1) is the first of n G.M. s between positive numbers a and b ,...

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  10. Find the 15^(th) term of the series 3 - 6 +9 - 12 + …

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  11. Find the r^(th) term of the series (2)/(4) + (4)/(4^(2)) + (6)/(4...

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  12. Find the following sums : (i) 1 + 2 + 3 + …+ 30 (ii) 1^(2) + 2^...

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  13. Find the following sums : (i) 5 + 6 + 7 +…+ 100 (ii) 15^(2) + 16...

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  14. Find the sum of n terms of the series whose nth terms is (i) n(n-1)...

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  15. Find the sum to n term of the series whose n^(th) term is given by ...

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  16. Find the sum of first n terms of the series 1^(3) + 3^(3) + 5^(3) +…

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  17. Find the sum of the series 1 . 3^(2) + 2.5 ^(2) + 3.7^(2) +…+ to n t...

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  18. Find the sum of n terms of series : 3 + 5 + 9 + 15 + 23 + ………… n terms

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  19. Find the n^(th) term of the series 2 + 5 + 10 + 17 + 26 + …

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  20. The sum to infinity of the series 1/2.4+1/4.6+1/6.8+1/8.10+....... is

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