Question

If 1+3+5+..... n terms/2+4+6..... 52 terms =2/51 then the value of n is​

Answer 1

Given.

1+3+5+..... n terms/2+4+6..... 52 terms =2/51

To find

The value of n

Explanation

Consider them as A.P 1 and A.P 2

Now,

A.P 1:

1,3,5,7.........n terms

$\boxed{\sf sum \: of \: n\: terms \:of\: A.P , S_n=\dfrac{n}{2}(2a+(n-1)d)}$

Here

a₁=a=1

common difference=d=a₂-a₁

d=3-1=2

Substituting in the formula

$S_1=\dfrac{n}{2}(2(1)+(n-1)2)$

$S_1=\dfrac{n}{2}(2+2n-2)$

$S_1=\dfrac{n}{2}\times2n$

$S_1=n^2.........(1)$

A.P 2:

2,4,6,.............52 terms

Here a¹=a=2

d=4-2=2

Substituting the values

$S_2=\dfrac{52}{2}(2(2)+(52-1)2)$

$S_2=26(4+102)$

$S_6=26(106)$

$S_6=2756.......(2)$

given

$\underline {\sf\dfrac{(1)}{(2)}=\dfrac{2}{51}}$

$\dfrac{ {n}^{2} }{2756} = \dfrac{2}{51}$

${n}^{2} = \dfrac{2756 \times 2}{51} \\ \\ = > n = 10 \: terms \:$

approximately

Answer 2

Answer: n = 10

Step-by-step explanation:

Given,

$\frac{1+3+5+................n\;terms}{2+4+6+................50\;terms}=\frac{2}{51}$

Considering first Arithmetic series in numerator:-

First Term = 1

Common Difference = 3 - 1 = 2

Number of terms = n

Let the sum of terms be S₁

We know that,

S₁ = n/2(2a + (n-1)d)

S₁ = n/2[(2×1) + 2(n-1)]

S₁ = n/2[2 + 2n - 2]

S₁ = n/2 × 2n

S₁ = n²

Considering first Arithmetic series in Denominator:-

First Term = 2

Common Difference = 4 - 2 = 2

Number of terms (n) = 50

Let the sum of terms be S₂

We know that,

S₂ = n/2(2a + (n-1)d)

S₂ = 50/2[(2×2) + 2(50 - 1)]

S₂ = 25[4 + 2*49]

S₂ = 25[4 + 98]

S₂ = 25 × 102

S₂ = 2550

So we have,

$\frac{S_1}{S_2}=\frac{2}{51}$

$\frac{n^2}{2550}=\frac{2}{51}\\\;\\n^2=\frac{2\times2550}{51}\\\;\\n^2=50\times2\\\;\\n^2=100\\\;\\n=10$