Question

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$if \: \frac{1 + 3 + 5... \: \: \: \: upto \: n}{2 + 5 + 8...upto \: 8} = 9$

Then find the value of n

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Answer 1

This is the answer to your question

Answer 2

(i) Numerator:

1 + 3 + 5 + 7....n.

Here, 1,3,5,7 is an arithmetic progression with:

⇒ First term a = 1.

⇒ Common difference d = a₂ - a₁ = 3 - 1 = 2.

We know that nth term of an AP an = a + (n - 1) * d

                                                           = 1 + (n - 1) * 2

                                                           = 1 + 2n - 2

                                                           = 2n - 1.

We know that Sum of n terms of an AP sn = (n/2)[a + l]

⇒ (n/2)[1 + 2n - 1]

⇒ 2n^2/2

⇒ n^2.

(ii) Denominator:

2 + 5 + 8 + 11....8.

Here, 2,5,8 is an Arithmetic Progression with:

⇒ First term a = 2.

⇒ Common difference d = a₂ - a₁ = 5 - 2 = 3.

We know that nth term of an AP an = a + (n - 1) * d

8th term of an AP a₈ = a + 7d

                                  = 2 + 7(3)

                                  = 23.

Sum of n terms of an AP sn = (n/2)[a + l]

                                              = (8/2)[2 + 23]

                                              = 4[25]

                                              = 100.

Now,

Given that (Numerator/Denominator) = 9.

⇒ (n^2)/100 = 9

⇒ n^2 = 900

⇒ n = √900

⇒ n = +30,-30.

Therefore, the value of n = 30.

Hope it helps!