Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
3.9k views
in Geometric Progressions by (24.0k points)
closed by

What is the sum of n terms of the series 0.2 + 0.22 + 0.222 + ......?

1 Answer

+1 vote
by (23.5k points)
selected by
 
Best answer

0.2 + 0.22 + 0.222 + ..... to n terms 

= 2[0.1 + 0.11 + 0.111 + ..... to n terms] 

= \(\frac{2}{9}\) [0.9 + 0.99 + 0.999 + ..... to n terms] 

= \(\frac{2}{9}\) [(1 – 0.1) + (1 – 0.01) + (1 + 0.001) + ..... to n terms]

= \(\frac{2}{9}\) [(1 + 1 + 1 + ..... to n terms) – (0.1 + 0.01 + 0.001 + ..... to n terms)]

\(\frac{2}{9}\)\(\bigg[n-\frac{0.1(1-(0.1)^n}{(1-0.1)}\bigg]\) = \(\frac{2}{9}\)\(\bigg[n-\frac{\frac{1}{10}\big(1-\frac{1}{10^n}\big)}{1-\frac{1}{10}}\bigg]\) = \(\frac{2}{9}\)\(\bigg[n-\frac{1}{9}\bigg(1-\frac{1}{10^n}\bigg)\bigg]\).

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...