Find the 10^{th} term of the A.P. sqrt{2}, 3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given $A.P.$ is $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, \ldots$Here,the first term $a = \sqrt{2}$.The common difference $d = 3 \sqrt{2} - \sqrt{2}

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(C) The given $A.P.$ is $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, \ldots$Here,the first term $a = \sqrt{2}$.The common difference $d = 3 \sqrt{2} - \sqrt{2} = 2 \sqrt{2}$.The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.For the $10^{th}$ term $(n = 10)$:$a_{10} = \sqrt{2} + (10 - 1)(2 \sqrt{2})$$a_{10} = \sqrt{2} + 9(2 \sqrt{2})$$a_{10} = \sqrt{2} + 18 \sqrt{2}$$a_{10} = 19 \sqrt{2}$.

Find the $10^{th}$ term of the $A.P.$ $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, \ldots$ (in $\sqrt{2}$)

Similar Questions

Find the sum of those integers between $1$ and $500$ which are multiples of $2$ as well as of $5$.

The sum of how many terms of the $A.P.$ $31, 36, 41, \ldots$ is $535$?

Which of the following sequences is an $A.P.$?

The $21^{\text{st}}$ term of the $AP$ whose first two terms are $-3$ and $4$ is

The sum of the first $n$ terms of an $A.P.$ is given by $S_{n} = 7n^{2} - 3n$. Find the $n^{th}$ term of the $A.P.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module