Question

please tell solution​

Answer 1

Answer:

Q is not in a proper way......

Answer 2

$\huge{\orange{\underline{\underline{Answer:-}}}}$

Given:

Terms in Arithmetic progression √2, 3√2, 5√2

To find:

10th term of AP = ?

Solution:

$\\ tn \: = a + (n - 1)d \\ \\where \: n \: = term \: to \: be \: find \\ a = first \: term \\ d = common \: \: difference$

Here, we can find common difference by subtracting second term with first term, then third term with second term. If the difference is same it becomes AP common difference (d)

$\\ 3 \sqrt{2} - \sqrt{2} = 2 \sqrt{2} \\ \\ 5\sqrt{2} - 3 \sqrt{2} = 2 \sqrt{2} \\ \\ common \: difference(d) = 2 \sqrt{2}$

Here, n = 10

d = 2√2

a = √2

$\\ tn \: = a + (n - 1)d \\ \\ \: substituting \: value \: in \:above \: eq {n}^{}$

$\\ \\ t10 = \sqrt{2} + (10 - 1)2 \sqrt{2 } \\ \\ = \sqrt{2} + 9 \times 2 \sqrt{2} \\ \\ = \sqrt{2} + 18 \sqrt{2} \\ \\ = 19 \sqrt{2}$

Required answer:

Therefore, 10th term of given Arithmetic progression √2, 3√2, 5√2 is 19√2

Basic points:

Solve more such questions to get good hold on it