If S1 is the sum of an arithmetic progression of ‘n’ odd number of terms and S2 is the sum of the terms of the series in odd places, then $$\frac{{{S_1}}}{{{S_2}}}$$
A. $$\frac{{2n}}{{n + 1}}$$
B. $$\frac{n}{{n + 1}}$$
C. $$\frac{{n + 1}}{{2n}}$$
D. $$\frac{{n - 1}}{n}$$
Answer: Option A
Solution (By Examveda Team)
Odd numbers are 1, 3, 5, 7, 9, 11, 13, ...... n∴ S1 = Sum of odd numbers = n2
S2 = Sum of number at odd places
3, 7, 11, 15, ......
a = 3, d = 7 - 3 = 4 and number of term = $$\frac{n}{2}$$
$$\eqalign{ & {S_2} = \frac{n}{{2 \times 2}}\left[ {2 \times 3 + \left( {\frac{n}{2} - 1} \right) \times 4} \right] \cr & \,\,\,\,\,\,\,\,\, = \frac{n}{4}\left[ {6 + 2n - 4} \right] \cr & \,\,\,\,\,\,\,\,\, = \frac{n}{4}\left[ {2n + 2} \right] \cr & \,\,\,\,\,\,\,\,\, = \frac{{n\left( {n + 1} \right)}}{2} \cr & \therefore \frac{{{s_1}}}{{{s_2}}} = \frac{{{n^2} \times 2}}{{n\left( {n + 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2n}}{{n + 1}} \cr} $$
This Question Belongs to Arithmetic Ability >> Progressions
Let the A.P. have a first term 'a' and a common difference 'd'.
The series is: a, a+d, a+2d, a+3d, ...
Step 1: Write the formula for S₁
S₁ is the sum of the entire A.P. with 'n' terms. This is a direct application of the standard sum formula:
Sₙ = (number of terms / 2) * [2a + (number of terms - 1)d]
For S₁, we have:
Number of terms = n
First term = a
Common difference = d
Plugging these in, we get:
S₁ = n/2 * [2a + (n-1)d] --- (Equation 1)
Step 2: Analyze and write the formula for S₂
S₂ is the sum of the terms in the odd places: 1st term, 3rd term, 5th term, ..., nth term.
This new series is also an A.P., but we need to find its specific properties.
First term of S₂: The first term is the same, which is a.
Common difference of S₂: The difference is between a₃ and a₁, then a₅ and a₃, and so on.
a₃ - a₁ = (a + 2d) - a = 2d
So, the new common difference is 2d.
Number of terms in S₂: Since 'n' is an odd number, the number of odd-placed terms from 1 to n is (n + 1) / 2.
(Example: if n=5, the terms are 1st, 3rd, 5th. That's 3 terms, and (5+1)/2 = 3).
Now, we apply the sum formula to S₂ using these new properties:
S₂ = (number of terms in S₂ / 2) * [2*(first term) + (number of terms in S₂ - 1)*(common difference)]
Substitute the properties we just found:
S₂ = ( ((n+1)/2) / 2 ) * [ 2a + ( ((n+1)/2) - 1 ) * (2d) ]
Let's simplify this expression for S₂:
The outer fraction becomes: (n+1) / 4
The part inside the bracket needs simplification: ((n+1)/2) - 1 = (n+1-2)/2 = (n-1)/2.
So the expression becomes: S₂ = (n+1)/4 * [ 2a + ( (n-1)/2 ) * (2d) ]
The 2s in the d term cancel out: S₂ = (n+1)/4 * [ 2a + (n-1)d ]
So, our final expression for S₂ is:
S₂ = (n+1)/4 * [2a + (n-1)d] --- (Equation 2)
Step 3: Calculate the Ratio S₁ / S₂
Now we divide Equation 1 by Equation 2:
S₁ / S₂ = { n/2 * [2a + (n-1)d] } / { (n+1)/4 * [2a + (n-1)d] }
Notice that the entire bracketed part [2a + (n-1)d] is identical in the numerator and the denominator. Therefore, it cancels out completely.
We are left with a simple fraction:
S₁ / S₂ = (n/2) / ((n+1)/4)
To divide by a fraction, you multiply by its reciprocal (flip it):
S₁ / S₂ = (n/2) * (4 / (n+1))
S₁ / S₂ = (4n) / (2(n+1))
Simplify by dividing the 4 by the 2:
S₁ / S₂ = 2n / (n+1)
This matches option A.
1+3+5+.......+n
Here, what do indicate odd places? Aren't the term numbers?
1 stays at 1st place,
5 at 3rd,
9 at 5th ....and so on.
aren't the odd term number indicating the odd places? So, why 3,7,11,15... are taken as odd place holders?