If the sum of the first n terms of a sequence | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We are given $S_n = An^2 + Bn$.$S_{n - 1} = A(n - 1)^2 + B(n - 1)$$= A(n^2 - 2n + 1) + B(n - 1)$$= An^2 - 2An + A + Bn - B$The $n^{th}$ term $a_n

Vedclass · Accepted Answer

Arithmetic Progression

Vedclass · Answer

(A) We are given $S_n = An^2 + Bn$.$S_{n - 1} = A(n - 1)^2 + B(n - 1)$$= A(n^2 - 2n + 1) + B(n - 1)$$= An^2 - 2An + A + Bn - B$The $n^{th}$ term $a_n = S_n - S_{n - 1} = (An^2 + Bn) - (An^2 - 2An + A + Bn - B)$$= An^2 + Bn - An^2 + 2An - A - Bn + B$$= 2An + B - A$Now,$a_{n - 1} = 2A(n - 1) + B - A = 2An - 2A + B - A = 2An + B - 3A$.The common difference $d = a_n - a_{n - 1} = (2An + B - A) - (2An + B - 3A) = 2A$.Since the difference $d = 2A$ is a constant independent of $n$,the sequence is an Arithmetic Progression.

If the sum of the first $n$ terms of a sequence is of the form $An^2 + Bn$,where $A$ and $B$ are constants independent of $n$,then the sequence is a ........

Similar Questions

If the sum of the first $n$ terms of an $A.P.$ is $cn(n - 1)$,where $c \neq 0$,then the sum of the squares of these terms is:

If $a$,$b$,and $c$ are positive real numbers,then $a/b + b/c + c/a$ is greater than or equal to what value?

If $a_1, a_2, a_3, \dots, a_{21}$ are in $A.P.$ and $a_3 + a_5 + a_{11} + a_{17} + a_{19} = 10$,then the value of $\sum_{r=1}^{21} a_r$ is:

Let $a_1, a_2, a_3, \ldots$ be in an arithmetic progression of positive terms. Let $A_{k}=a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_{2k-1}^2-a_{2k}^2$. If $A_3=-153$,$A_5=-435$ and $a_1^2+a_2^2+a_3^2=66$,then $a_{17}-A_7$ is equal to:

If the sum of $n$ terms of an $A.P.$ is $nA + n^2B$,where $A$ and $B$ are constants,then its common difference will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module