Question

If in a series $S_{n}=a n^{2}+b n+c$, where $S_{n}$ denotes the sum of $n$ terms, then

• A. The series is always arithmetic
• B. The series is arithmetic from the second term onwards
• C. The series may or may not be arithmetic
• D. The series cannot be arithmetic

Answer 1

Answer: (B)

$S_{n}=a n^{2}+b n+c$
$\therefore S_{n-1}=a(n-1)^{2}+b(n-1)+c$ for $n \geq 2$
$\therefore t_{n}=S_{n}-S_{n-1}$
$=a\left\{n^{2}-(n-1)^{2}\right\}+b\{n-(n-1)\}=a(2 n-1)+b$
$\therefore t_{n}=2 an+b-a, n \geq 2$
$\therefore t_{n-1}=2a(n-1)+b-a$ for $n \geq 3$
$\therefore t_{n}-t_{n-1}=2a(n-n+1)=2a$ for $n \geq 3$
$\therefore t_{3}-t_{2}=t_{4}-t_{3}= ..........2a$
Now $t_{2}-t_{1}=\left(S_{2}-S_{1}\right)-S_{1}=S_{2}-2 S_{1}$
$=\left(a \cdot 2^{2}+b \cdot 2+c\right)-\left\{a \cdot 1^{2}+b \cdot 1+c\right\}=2 a-c \neq 2 a$
$\therefore $ Series is arithmetic from the second term onwards.