- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Show that $a_1, a_2, ……. a_n,……$ form $a_n$ AP where an is defined as below:
$a_n = 9 - 5n$
Also find the sum of the first 15 terms in each case.
Given:
$a_n=9-5n$
To do:
We have to show that $a_1, a_2, ……. a_n,……$ forms an AP and find the sum of the first 15 terms.
Solution:
To find the given sequence we have to substitute $n=1, 2, 3.....$ in $a_n=9-5n$.
Therefore,
$a_1=9-5(1)$
$=9-5$
$=4$
$a_2=9-5(2)$
$=9-10$
$=-1$
$a_3=9-5(3)$
$=9-15$
$=-6$
$a_4=9-5(4)$
$=9-20$
$=-11$
The sequence formed is $4, -1, -6, -11,.....$.
For the given sequence to form an A.P., the difference between any two consecutive terms should be equal.
Here,
$d=a_2-a_1=-1-4=-5$
$d=a_3-a_2=-6-(-1)=-6+1=-5$
$d=a_4-a_3=-11-(-6)=-11+6=-5$
This implies,
$a_2-a_1=a_3-a_2=a_4-a_3=d$
Therefore, the given sequence forms an A.P. 
 We know that,$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$S_{15}=\frac{15}{2}[2 a+(15-1) d]$
$=\frac{15}{2}[2 \times 4+(15-1) \times (-5)]$
$=\frac{15}{2}[8+14 \times (-5)]$
$=\frac{15}{2}[8-70]$
$=\frac{15}{2} \times (-62)$
$=15 \times (-31)$
$=-465$
The sum of the first 15 terms is $-465$.