Question

If a$_1$ = 4 and $a_{n+1}=a_{n}+4n\quad for\quad n\ge1. $ then the value of a$_{100}$ is

• A. 19804
• B. 18904
• C. 18894
• D. 19904
• E. 19894

Answer 1

Answer: (A)

Given, $a_{n+1}=a_{n}+4 n$ and $a_{1}=4$
$\Rightarrow \, a_{n+1}-a_{n}=4 n$
Put $n=1, \, a_{2}-a_{1}=4 \cdot 1$
Put $n=2 \, a_{3}-a_{2}=4 \cdot 2$
Put $n=3, \, a_{4}-a_{3}=4 \cdot 3$
$…… \,\,\,\,\,\,\,\,\, …… \,\,\,\,\,\,\,\,\, …… \,\,\,\,\,\,\,\,\, $
$…… \,\,\,\,\,\,\,\,\, …… \,\,\,\,\,\,\,\,\, …… \,\,\,\,\,\,\,\,\, $
Put $n=99,\, a_{100}-a_{99}=4 \cdot 99$
On adding, we get
$a_{100}-a_{1}=4(1+2+\ldots+99)$
$\Rightarrow \, a_{100}-4=4 \cdot \frac{99(99+1)}{2}$
$\left[\because \sum n=\frac{n(n+1)}{2}\right]$
$\therefore \, a_{100}=19804$