# Sequences and Series - Class 11 - Mathematics

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## Extra Questions - Sequences and Series | NCERT | Mathematics | Class 11

If $x, y, z$ are the $15^{\text{th}}, 20^{\text{th}}, 25^{\text{th}}$ terms of a G.P., then find the value of

$$ \Delta = \begin{bmatrix} \ln x & 15 & 1 \\ \ln y & 20 & 1 \\ \ln z & 25 & 1\end{bmatrix} $$

A) 0 B) 2 C) 1 D) 4

To solve the problem, we start by recognizing that $x, y, z$ are the 15th, 20th, and 25th terms of a geometric progression (G.P.). Let's denote the first term of the G.P. as $a$ and the common ratio as $r$. Then the $n$-th term of a G.P. is given by $ar^{n-1}$. Thus, we can write:

- $x = ar^{15-1} = ar^{14}$

- $y = ar^{20-1} = ar^{19}$

- $z = ar^{25-1} = ar^{24}$

We need to evaluate the determinant of the matrix:

$$ \Delta = \begin{bmatrix} \ln x & 15 & 1 \\ \ln y & 20 & 1 \\ \ln z & 25 & 1\end{bmatrix} $$

Plugging in the expressions for $\ln x = \ln (ar^{14})$, $\ln y = \ln (ar^{19})$, $\ln z = \ln (ar^{24})$, we can simplify using logarithmic properties:

- $\ln x = \ln a + 14 \ln r$

- $\ln y = \ln a + 19 \ln r$

- $\ln z = \ln a + 24 \ln r$

Thus,

$$ \Delta = \begin{bmatrix} \ln a + 14 \ln r & 15 & 1 \\ \ln a + 19 \ln r & 20 & 1 \\ \ln a + 24 \ln r & 25 & 1\end{bmatrix} $$

Recall that if we factor out any common element in a row or a column of a determinant, it simplifies the calculation. We can subtract a multiple of one row from another to simplify further. Let's operate on the first column:

- Subtract $\ln a$ (this is factored out and doesn't change the determinant's value).

- We then have $\ln r$ as a common factor in the first column which we can factor out:

$$ \Delta = \ln r \cdot \begin{bmatrix} 14 & 15 & 1 \\ 19 & 20 & 1 \\ 24 & 25 & 1 \end{bmatrix} $$

Now in this new matrix, each entry in the first column is a simple arithmetic sequence with a common difference. Thus we proceed:

- Apply row operations: subtract Row 1 from Row 2, and subtract Row 2 from Row 3.

- This gives:

$$ = \ln r \cdot \begin{bmatrix} 14 & 15 & 1 \\ 19-14 & 20-15 & 1-1 \\ 24-19 & 25-20 & 1-1 \end{bmatrix} $$

$$ = \ln r \cdot \begin{bmatrix} 14 & 15 & 1 \\ 5 & 5 & 0 \\ 5 & 5 & 0 \end{bmatrix} $$

Since the last two rows are identical, the determinant now becomes zero due to linearly dependent rows in a determinant calculation. Thus,

$$ \Delta = 0. $$

The answer is:

A) 0.

Find the missing term:

5, 9, 16, 29, 54, 103, (......)

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If the roots of the equation are equal, then the discriminant must be equal to 0. The discriminant of a quadratic equation $ax^{2}+bx+c=0$ is given by $b^{2}-4ac$.

For the given equation, $a(b-c)$, $b(c-a)$, and $c(a-b)$ are the coefficients of $x^{2}$, $x$, and the constant term respectively.

Given that the roots are equal, the discriminant can be written as: $(b(c-a))^{2}-4(a(b-c))(c(a-b))=0$

Expanding the above equation: $b^{2}c^{2}-2abc^{2}+a^{2}b^{2}+4abc^{2}-4a^{2}bc=0$

Simplifying: $b^{2}c^{2}+a^{2}b^{2}-2abc^{2}-4a^{2}bc=0$

$a^{2}b^{2}+b^{2}c^{2}-2abc(c+a)=0$

$a^{2}b^{2}+b^{2}c^{2}-2abc^{2}-2ab^{2}c=0$

Factoring out common terms: $a^{2}b^{2}+b^{2}c^{2}-2abc(c+b)=0$

$a^{2}b^{2}+b^{2}c^{2}-2ab(c^{2}+bc)=0$

$a^{2}b^{2}+b^{2}c^{2}=2ab(c^{2}+bc)$

Dividing by $a^{2}b^{2}c^{2}$, we get: $1/a^{2}+1/b^{2}=2/(c^{2}+bc)$

This can be written as: $1/a^{2}+1/b^{2}=2/(c(c+b))$

Now, the reciprocals of $a$ and $b$ are in AP with the common difference $2/(c(c+b))$. Therefore, $a$, $b$, and $c$ are in HP.

Therefore, the correct option is: Assertion is true and Reason explains the assertion.