Question

Algebraic expression of sum of terms of an arithmetic sequence is 3n2+4n. (a) Find the first term?
(b) What is the algebraic expression of the sequence?
(c) Fin the 20th term?

Answer 1

Step-by-step explanation:

• the first term is 3n^2
• the algebraic expression is quadratic
• sorry I know only these two

Answer 2

The first term is 13.The algebraic expression is 6n+7.The 20th term is 127.

Step-by-step explanation:

Given ,$S_{n}$=$3n^{2}+4n$

• The first term=$S_{2} -S_{1}$.

⇒$3\times(2)^{2}+4\times(2)-[3(1)^{2}+4]$

⇒12+8-3-4

⇒13          

• Algebraic expression=$S_{n+1} -S_{n}$

⇒$3\times(n+1)^{2}+4(n+1)-3(n^{2})-4n$

⇒$6n+7$

• 20th Term=$S_{21} -S_{20}$

⇒$3\timess(21)^{2}+4\times(21)-[3\times(20)^{2}+4(20)]$

⇒$3\times441+4\times21-3\times400-4\times20$

⇒$1323+84-1200-80$

⇒127