Question

The expression for the sum to 'n' terms of some Arithmetic Sequence are given beliw. Find the 'nth' term of each

i)n^2+2n
ii) n^2- 2n​

Answer 1

Answers:-

1) Given:

Sum of first n terms = n² + 2n

We know that,

Sum of first n terms = n/2 * [ a + l ]

Where,

• a is first term
• l is nth or last term.

Substitute n = 1 to find the sum of first 1 terms i.e., first term (a) of the AP.

→ S₁ = a = (1)² + 2(1)

→ a = 1 + 2

→ a = 3

Hence,

→ n/2 * [ a + l ] = n² + 2n

→ a + l = n(n + 2) * 2/n

→ a + l = 2n + 4

→ l = 2n + 4 - a

→ l = 2n + 4 - 3

→ l = 2n + 1

Hence, the nth term is 2n + 1.

2) Given:

S(n) = n² - 2n

→ S₁ = a = (1)² - 2(1)

→ a = - 1

Hence,

→ n/2 * [ - 1 + l ] = n² - 2n

→ l - 1 = n(n - 2) * 2/n

→ l - 1 = 2n - 4

→ l = 2n - 4 + 1

→ l = 2n - 3

Hence, the nth term is 2n - 3.

Answer 2

$\bf{\gray{\underbrace{\blue{GIVEN:-}}}}$

• The sum of nth term,

1. n² + 2n
2. n² - 2n

$\bf{\gray{\underbrace{\blue{TO\:FIND:-}}}}$

• The nth term .

$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}}$

✨ The form of " $\bf{S_n}$" in the A.P is,

$\huge\purple\checkmark$ $\bf\green{S_n\:=\:\dfrac{n}{2}\:[a_1\:+\:a_n]\:}$

Where,

• $\bf{a_1}$ = First term of A.P .

• $\bf{a_n}$ = nth term of A.P .

$\rule{200}2$

[1]

$\bf\begin{tabular}{| c | c | c |}n & S_n = n^2 + 2n & Series \\ \cline{1 - 3}1 & S_1 = 3 & a_1 = 3 \\ \cline{1 - 3}2 & S_2 = 8 & a_2 = 8 - 3 = 5 \\ \cline{1 - 3}3 & S_3 = 15 & a_3 = 15 - 8 = 7 \\ \cline{1 - 3}4 & S_4 = 24 & a_4 = 24 - 15 = 9 \\ \cline{1 - 3}\end{tabular}$

$\orange\bigstar$ $\bf\pink{A.P\:=\:3\:,\:5\:,\:7\:,\:9\:,\:....\:}$

$\bf{\implies\:S_n\:=\:n^2\:+\:2n\:}$

[Note :- Taking common $\bf{\dfrac{n}{2}}$ from the above equation .]

$\rm{\implies\:S_n\:=\:\dfrac{n}{2}\:[2n\:+\:4]\:}$

$\rm{\implies\:S_n\:=\:\dfrac{n}{2}\:[{\red{\underline{3}}}\:+\:{\red{\underline{2n\:+\:1}}}]\:}$

✨ Hence,

• $\bf{a_1\:=\:3}$

• $\bf\purple{nth\:term\:=\:2n\:+\:1}$

$\red\therefore$ The nth term is "2n + 1" .

$\rule{200}2$

[2]

$\bf\begin{tabular}{| c | c | c |}n & S_n = n^2 - 2n & Series \\ \cline{1 - 3}1 & S_1 = -1 & a_1 = -1 \\ \cline{1 - 3}2 & S_2 = 0 & a_2 = 0 - (-1) = 1 \\ \cline{1 - 3}3 & S_3 = 3 & a_3 = 3 - 0 = 3 \\ \cline{1 - 3}4 & S_4 = 8 & a_4 = 8 - 3 = 5 \\ \cline{1 - 3}\end{tabular}$

$\green\bigstar$ $\bf\pink{A.P\:=\:-1\:,\:1\:,\:3\:,\:5\:,\:....\:}$

$\bf{\implies\:S_n\:=\:n^2\:-\:2n\:}$

[Note :- Taking common $\bf{\dfrac{n}{2}}$ from the above equation .]

$\rm{\implies\:S_n\:=\:\dfrac{n}{2}\:[2n\:-\:4]\:}$

$\rm{\implies\:S_n\:=\:\dfrac{n}{2}\:[{\red{\underline{-1}}}\:+\:{\red{\underline{2n\:-\:3}}}]\:}$

✨ Hence,

• $\bf{a_1\:=\:-1}$

• $\bf\purple{nth\:term\:=\:2n\:-\:3}$

$\red\therefore$ The nth term is "2n - 3" .