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Geometric Progression
1. For given geometric progression series 3,6,12,24,48 ,... find 10 th term and addition of first 10 th terms.
2. For given geometric progression series 3,6,12,24,48 ,... then find n such that S(n) = 3069 .
3. For given geometric progression series 3,6,12,24,48 ,... then find n such that f(n) = 1536 .
4. For geometric progression f( 1 ) = 2 , f( 4 ) = 54 then find f( 3 ) and S( 3 ).
5. For geometric progression f( 1 ) = 2 , f( 4 ) = 54 , then find n such that f(n) = 18 .
6. For geometric progression addition of 3 terms is 26 and their multiplication is 216 , then that numbers
7. For geometric progression multiplication of 5 terms is 1 and 5 th term is 81 times then the 1 th term.
8. Arithmetic mean of two number is 13 and geometric mean is 12 , then find that numbers
9. Two numbers are in the ratio 9 : 16 and difference of arithmetic mean and geometric mean is 1 , then find that numbers
10. Find 6 arithmetic mean between 3 and 24 .
11. Find 3 geometric mean between 1 and 256 .
12. Prove that 1 + (1 + 2) + (1 + 2 + 3) + ... n terms = n/6 (n + 1) (n + 2)
13. Prove that 1 * (2^2 - 3^2) + 2 * (3^2 - 4^2) + 3 * (4^2 - 5^2) + ... n terms = n/6 (n + 1) (4n + 11)
14. 1 + x^4 + 3^2 + 4 + x^6 + 6^2 + 7 + x^8 + 9^2 + ... 3n terms
15. 1 + (1 + 3) + (1 + 3 + 5) + ... n terms
16. Prove that for all n belongs to N, 1^2 * n + 2^2 * (n - 1) + 3^2 * (n - 2) + ... + n^2 * 1 = n/12 (n + 1)^2 (n + 2)
17. Prove that 1 * 2^2 + 3 * 5^2 + 5 * 8^2 + ... n terms = n/2 (9n^3 + 4n^2 - 4n - 1)
18. Prove that 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... n terms = n/12 (n + 1)^2 (n + 2)
19. Prove that 2 + 5 + 10 + 17 + ... n terms = n/6 (2n^2 + 3n + 7)
20. Prove that sum [ sum (2n -3) ] = n/6 (n + 1)(2n - 5)
21. For geometric progression, find 1 + 1/sqrt(2) + 1/2 + 1/(2*sqrt(2)) + ... 10 terms ( For geometric progression, find 1 + ...
22. Find 1^2 + 2^2 + ...+ 10^2 , ( Find a^2 + b^2 + ... + n^2 , where a = 1 , b = 2 and n = 10 . )
23. Find 1^2 + 2^2 + ... 10 terms, ( Find a^2 + b^2 + ... n terms, where a = 1 , b = 2 and n = 10 . )
Problem 20 of 23
20. Prove that `sum [ sum (2n -3) ] = n/6 (n + 1)(2n - 5)`
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Geometric Progression
20. Prove that `sum [ sum (2n -3) ] = n/6 (n + 1)(2n - 5)`
L.H.S. `= sum [ sum (2n -3) ]`
`= sum [ sum 2n - sum 3 ]`
`= sum [ 2 × n/2 (n + 1) -3n ]`
`= sum [ n (n + 1) - 3n ]`
`= sum [ n^2 + n - 3n ]`
`= sum [ n^2 - 2n ]`
`= sum n^2 - 2 sum n`
`= (n (n + 1) (2n + 1))/6 - 2 * (n (n + 1))/2`
`= (n (n + 1) [ (2n + 1) - 6 ])/6`
`= (n (n + 1) (2n - 5))/6`
`=` R.H.S. (Proved)
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