Question

if the sum of 'N' terms of an AP is 3n2
-2n. find the nth term of that AP​

Answer 1

EXPLANATION.

Sum of n terms of an A.P. = 3n² - 2n.

As we know that,

⇒ Tₙ = Sₙ - Sₙ₋₁.

Using this formula in equation, we get.

⇒ 3n² - 2n - [3(n - 1)² - 2(n - 1)].

⇒ 3n² - 2n - [3(n² + 1 - 2n) - 2n + 2].

⇒ 3n² - 2n - [3n² + 3 - 6n - 2n + 2].

⇒ 3n² - 2n - [3n² - 8n + 5].

⇒ 3n² - 2n - 3n² + 8n - 5.

⇒ - 2n + 8n - 5.

⇒ 6n - 5. = Algebraic expression.

Put the value of n = 1 in equation, we get.

⇒ 6(1) - 5.

⇒ 6 - 5.

⇒ 1.

Put the value of n = 2 in equation, we get.

⇒ 6(2) - 5.

⇒ 12 - 5.

⇒ 7.

Put the value of n = 3 in equation, we get.

⇒ 6(3) - 5.

⇒ 18 - 5.

⇒ 13.

Put the value of n = 4 in equation, we get.

⇒ 6(4) - 5.

⇒ 24 - 5.

⇒ 19.

Series = 1, 7, 13, 19,,,,,,,

First term = a = 1.

Common difference = d = b - a = c - b.

Common difference = d = 7 - 1 = 13 - 7.

Common difference = d = 6.

As we know that,

General term of an A.P.

⇒ Tₙ = a + (n - 1)d.

⇒ Tₙ = 1 + (n - 1)6.

⇒ Tₙ = 1 + (6n - 6).

⇒ Tₙ = 1 + 6n - 6.

⇒ Tₙ = 6n - 5.

                                                                                                                         

MORE INFORMATION.

Relation between A.M. & G.M. & H.M.

If A, G, H are A.M , G.M , H.M between any two numbers.

(1) = A ≥ G ≥ H (equality holds when all terms are equal ).

(2) = G² = AH.

(3) = If A and G are A.M , G.M respectively between two positive numbers, then these numbers are,

A + √A² - G² , A - √A² - G².

Answer 2

Question:-

• If the sum of 'N' terms of an AP is 3n²  - 2n. Find the nth term of that AP.

Solution:-

$\tt\implies \: { 3n }^{ 2 } - 2n - [ 3{ ( n - 1 ) }^{ 2 } - 2( n - 1 ) ]$

$\tt\implies \: { 3n }^{ 2 } - 2n - [ 3({ n }^{ 2 } + 1 - 2n ) - 2( n - 1 ) ]$

$\tt\implies \: { 3n }^{ 2 } - 2n - [ { 3n }^{ 2 } + 3 - 6n - 2n + 2 ]$

$\tt\implies \: { 3n }^{ 2 } - 2n - [ { 3n }^{ 2 } - 8n + 5 ]$

$\tt\implies \: { 3n }^{ 2 } - 2n - { 3n }^{ 2 } + 8n - 5$

$\tt\implies \: 6n - 5$

Now ,

Take n = 1 in equation , then we get:-

$\tt\implies \: 6( 1 ) - 5$

$\tt\implies \: 6 - 5$

$\tt\implies \: 1$

Take n = 2 in equation , then we get:-

$\tt\implies \: 6( 2 ) - 5$

$\tt\implies \: 12 - 5$

$\tt\implies \: 7$

• Series = 1 , 7 , 13 , 19

Common difference = b - a

Common difference = 7 - 1

Common difference = 6

Arithmetic Progression formula

Tn = a + ( n - 1 )d

Tn = 1 + ( n - 1 )6

Tn = 1 + 6n - 6

Tn = 6n - 5