What does the series 1 + 3^{-1/2} + 3 + rac{1}{3 ext{√}3} + ... represent? (a) AP (b) GP (c) HP (d) None. What is the sum of the series 1 - rac{1}{2} + rac{1}{2} - rac{1}{8} +... What does the series 1 + 3^{-1/2} + 3 + rac{1}{3 ext{√}3} + ... represent? (a) AP (b) GP (c) HP (d) None. What is the sum of the series 1 - rac{1}{2} + rac{1}{2} - rac{1}{8} + ... ? (a) 1/2 (b) 3/2 (c) 2. If 1/4, 1/x and 1/10 are in HP, then what is the value of x? Read more

Steps to Solve

1. Identify the type of series
   We have the series ( 1 + 3^{-1/2} + 3 + \frac{1}{3\sqrt{3}} + \ldots ).
   To determine if it's an Arithmetic Progression (AP), Geometric Progression (GP), or Harmonic Progression (HP), we examine the ratios or differences of consecutive terms.

2. Check for Geometric Progression (GP)
   Calculate the ratios of consecutive terms:
   First terms:

• (3^{-1/2} / 1 = \frac{1}{\sqrt{3}})
• (3 / 3^{-1/2} = 3 \sqrt{3})
• (\frac{1}{3\sqrt{3}} / 3 = \frac{1}{9})

Since the ratios are not constant, it's not a GP.

3. Check for Harmonic Progression (HP)
   To check for HP, we consider the reciprocals of the terms:
   1, ( \sqrt{3} ), ( \frac{1}{3} ), ( 3 \sqrt{3} ).
   Calculating differences for AP in these reciprocals does not yield constant results, so it’s also not HP.

4. Conclude the series type
   Since the series is neither AP nor GP nor HP, it falls into the category of "None of the above" in this case.

5. Calculate the sum of the second series
   The second series is ( 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{8} + \ldots ).
   This series can be written as:
   ( S = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{8} + \ldots )

6. Identify and use formula for sums
   The absolute value of the negative terms forms a geometric series with first term ( \frac{1}{2} ) and common ratio ( -\frac{1}{2} ).
   The sum of an infinite geometric series is given by ( \frac{a}{1 - r} ), where ( a ) is the first term and ( r ) is the common ratio.

7. Calculate the geometric series sum
   Now calculating the sum of the series: $$ a = \frac{1}{2}, r = -\frac{1}{2} $$
   Thus the sum ( S ) is:

$$ S = \frac{\frac{1}{2}}{1 - (-\frac{1}{2})} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3} $$

However, since the series signs alternate, sum total is ( 1 - \text{Sum of series} ).

8. Final calculations
   Calculating:

• Starting from 1, subtracting alternating sums results in: $$ S = 1 - \left( \frac{1/2 + 1/8}{1 - (-1/2)} \right) = 1 - \frac{1/2 + 1/8}{3/2} = 1 - \frac{2 + 1}{3} = 1 - 1 = 0 $$
  So we will review and find the total.