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If three numbers are in H.P., then the n...

If three numbers are in H.P., then the numbers obtained by subtracting half of the middle number from each of them are in

A

A.P.

B

G.P.

C

H.P.

D

none of these

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To solve the problem step by step, we need to analyze the situation where three numbers are in Harmonic Progression (H.P.) and then determine the nature of the numbers obtained by subtracting half of the middle number from each of them. ### Step-by-Step Solution: 1. **Understanding H.P.**: - If three numbers \( A, B, C \) are in H.P., it means that their reciprocals \( \frac{1}{A}, \frac{1}{B}, \frac{1}{C} \) are in Arithmetic Progression (A.P.). - Therefore, we have the relation: \[ \frac{1}{A} + \frac{1}{C} = 2 \cdot \frac{1}{B} \] 2. **Rearranging the H.P. condition**: - From the H.P. condition, we can rearrange it to: \[ \frac{A + C}{AC} = \frac{2}{B} \] - This implies: \[ B = \frac{2AC}{A + C} \] 3. **Subtracting half of the middle number**: - We need to find the numbers obtained by subtracting half of the middle number \( B \) from each of the numbers \( A, B, C \). - Half of \( B \) is \( \frac{B}{2} \). - The new numbers will be: \[ A - \frac{B}{2}, \quad B - \frac{B}{2}, \quad C - \frac{B}{2} \] - This simplifies to: \[ A - \frac{B}{2}, \quad \frac{B}{2}, \quad C - \frac{B}{2} \] 4. **Finding the product of the first and last terms**: - We will now check if the new numbers are in Geometric Progression (G.P.) by examining the product of the first and last terms: \[ (A - \frac{B}{2})(C - \frac{B}{2}) \] 5. **Expanding the product**: - Expanding this gives: \[ AC - \frac{AB}{2} - \frac{BC}{2} + \frac{B^2}{4} \] 6. **Substituting \( B \)**: - We substitute \( B = \frac{2AC}{A + C} \) into the expression: \[ AC - \frac{1}{2} \cdot \frac{2AC}{A + C} \cdot A - \frac{1}{2} \cdot \frac{2AC}{A + C} \cdot C + \frac{1}{4} \cdot \left(\frac{2AC}{A + C}\right)^2 \] 7. **Simplifying**: - After simplification, we find that the product simplifies to: \[ \left(\frac{B}{2}\right)^2 \] - This shows that the new numbers are in G.P. because the product of the first and last terms equals the square of the middle term. ### Conclusion: Thus, the numbers obtained by subtracting half of the middle number from each of the three numbers in H.P. are in **Geometric Progression (G.P.)**.
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OBJECTIVE RD SHARMA ENGLISH-SEQUENCES AND SERIES-Chapter Test
  1. Let a,b,c be three positive prime number. The progrrssion in which sqr...

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  2. If 1/(b-a)+1/(b-c)=1/a+1/c , then (A). a ,b ,a n dc are in H.P. (B). a...

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  3. If three numbers are in H.P., then the numbers obtained by subtracting...

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  4. The first three of four given numbers are in G.P. and their last three...

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  5. In a G.P. of positive terms if any terms is equal to the sum of next ...

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  6. If a,b,c are in H.P and ab+bc+ca=15 then ca=

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  7. If sum(r=1)^(oo)(1)/((2r-1)^(2))=(pi^(2))/(8), then sum(r=1)^(oo) (1)/...

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  8. It is given that 1/1^4 + 1/2^4 +1/3^4 … to oo= pi^4/90 , then 1/1^4...

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  9. The minimum number of terms from the beginning of the series 20+22(2)/...

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  10. The sum of the series 1-3+5-7+9-11+ . . . . To n terms is

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  11. If three positive unequal numbers a, b, c are in H.P., then

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  12. If the fifth term of a G.P. is 2, then write the product of its 9 t...

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  13. 1^3-2^3+3^3-4^3+........+9^3 is equal to

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  14. The sum of infinite number of terms in G.P. is 20 and the sum of their...

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  15. If 1, log (9) (3^(1 - x) + 2) and log(3) (4.3^(x) -1) are A.P. then...

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  16. Two sequences lta(n)gtandltb(n)gt are defined by a(n)=log((5^(n+1))/...

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  17. The sum of the series (1)/(sqrt(1)+sqrt(2))+(1)/(sqrt(2)+sqrt(3))+(1...

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  18. Natural numbers are written as 1, (2,3), (4,5,6).. Show that the sum...

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  19. If the first term of an A.P. is 2 and common difference is 4, then ...

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  20. If 1+(1+2)/2+(1+2+3)/3+ddotto\ n terms is Sdot Then, S is equal to (n(...

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