Test Index

NCERT Class XI Mathematics - Sequences and Series - Solutions

© examsnet.com
Question : 42 of 106
Marks: +1, -0
7,21,37\sqrt{7},\sqrt{21},3\sqrt{7} , ... n terms.
Solution:  
In the given G.P.
a = 7\sqrt{7} , r = 3\sqrt{3} , sns_n = a(rn−1)r−1\frac{a(r^n-1)}{r-1} , if |r| > 1
∴ SnS_n = 7[(3)n−1]3−1\frac{\sqrt{7}[(\sqrt{3})^n-1]}{\sqrt{3}-1} = 73−1×3+13+1[(3)n−1]\frac{\sqrt{7}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}[(\sqrt{3})^n-1]
= 7(3+1)3−1[(3)n−1]\frac{\sqrt{7}(\sqrt{3}+1)}{3-1}[(\sqrt{3})^n-1]
∴ SnS_n = 72(3+1)(3n/2−1)\frac{\sqrt{7}}{2}(\sqrt{3}+1)(3^{n/2}-1)
© examsnet.com
Go to Question:
Prev Question
Next Question