First slide

Arithmetic progression

Question

$The sum of the first n-terms of the series \sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots is 435 \sqrt{3} then n is$

Moderate

A

18
B

15
C

13
D

29

Solution

$\begin{array}{l} given that \sqrt{3} [1 + 5 + 9 + 13 + \dots + T_{n}] = 435 \sqrt{3} \\ \Rightarrow \frac{n}{2} [2 + 4 (n - 1)] = 435 \\ \Rightarrow 2 n^{2} - n - 435 = 0 \end{array}$
n=15

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App