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S(n) be the sum of n terms of the series...

`S_(n)` be the sum of n terms of the series `(8)/(5)+(16)/(65)+(24)/(325)+"......"`
The seveth term of the series is (a)`(56)/(2505)` (b)`(56)/(6505)` (c)`(56)/(5185)` (d)`(107)/(9605)`

A

`(56)/(2505)`

B

`(56)/(6505)`

C

`(56)/(5185)`

D

`(107)/(9605)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the seventh term of the series given by \( S_n = \frac{8}{5} + \frac{16}{65} + \frac{24}{325} + \ldots \), we need to identify the general term \( T_n \) of the series. ### Step-by-Step Solution: 1. **Identify the General Term**: The series appears to have a pattern in both the numerators and denominators. By examining the first few terms: - The numerators are \( 8, 16, 24, \ldots \), which can be expressed as \( 8n \) where \( n \) is the term number. - The denominators are \( 5, 65, 325, \ldots \). We can see that these denominators can be expressed as \( 4n^4 + 1 \). Thus, we can write the \( n \)-th term as: \[ T_n = \frac{8n}{4n^4 + 1} \] 2. **Calculate the Seventh Term**: Now, we need to find \( T_7 \): \[ T_7 = \frac{8 \times 7}{4 \times 7^4 + 1} \] First, calculate \( 7^4 \): \[ 7^4 = 2401 \] Now substitute this value into the denominator: \[ 4 \times 7^4 + 1 = 4 \times 2401 + 1 = 9604 + 1 = 9605 \] Now substitute back into the expression for \( T_7 \): \[ T_7 = \frac{56}{9605} \] 3. **Final Result**: Thus, the seventh term of the series is: \[ T_7 = \frac{56}{9605} \] ### Answer: The correct option is (d) \( \frac{56}{9605} \).

To find the seventh term of the series given by \( S_n = \frac{8}{5} + \frac{16}{65} + \frac{24}{325} + \ldots \), we need to identify the general term \( T_n \) of the series. ### Step-by-Step Solution: 1. **Identify the General Term**: The series appears to have a pattern in both the numerators and denominators. By examining the first few terms: - The numerators are \( 8, 16, 24, \ldots \), which can be expressed as \( 8n \) where \( n \) is the term number. - The denominators are \( 5, 65, 325, \ldots \). We can see that these denominators can be expressed as \( 4n^4 + 1 \). ...
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ARIHANT MATHS ENGLISH-SEQUENCES AND SERIES-Exercise (Passage Based Questions)
  1. S(n) be the sum of n terms of the series (8)/(5)+(16)/(65)+(24)/(325)+...

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  2. S(n) be the sum of n terms of the series (8)/(5)+(16)/(65)+(24)/(325)+...

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  3. S(n) be the sum of n terms of the series (8)/(5)+(16)/(65)+(24)/(325)+...

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  4. Two consecutive numbers from 1,2,3 …., n are removed .The arithmetic m...

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  5. Two consecutive numbers from 1,2,3 …., n are removed .The arithmetic m...

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  6. Two consecutive numbers from 1,2,3 …., n are removed .The arithmetic m...

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  7. There are two sets A and B each of which consists of three numbers in ...

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  8. There are two sets A and B each of which consists of three numbers in ...

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  9. There are two sets A and B each of which consists of three numbers in ...

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  10. There are two sets A and B each of which consists of three numbers in ...

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  11. There are two sets A and B each of which consists of three numbers in ...

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  12. There are two sets A and B each of which consists of three numbers in ...

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  13. The numbers 1,3,6,10,15,21,28"..." are called triangular numbers. Let...

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  14. The numbers 1,3,6,10,15,21,28"..." are called triangular numbers. Let...

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  15. The numbers 1,3,6,10,15,21,28"..." are called triangular numbers. Let...

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  16. Let A(1),A(2),A(3),"......."A(m) be arithmetic means between -3 and 82...

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  17. Let A(1),A(2),A(3),"......."A(m) be arithmetic means between -3 and 82...

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  18. Let A(1),A(2),A(3),"......."A(m) be arithmetic means between -3 and 82...

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  19. Suppose alpha, beta are roots of ax^(2)+bx+c=0 and gamma, delta are ro...

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  20. Suppose alpha, beta are roots of ax^(2)+bx+c=0 and gamma, delta are ro...

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