Question

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Answer 1

Question -

Find the first term , the common difference and $t_{n}$ for a series having $\sf{ S_{n} = 3 {n}^2 + 5n }$

Solution -

We have the following given series -

$\sf{ S_{n} = 3 {n}^2 + 5n } \\ \\ \sf{ First \ Term \ - } \\ \\ \sf{ S_{1} = 3{1}^2 + 5 = 3 + 5 = 8 } \\ \\ \sf{ Hence , t_{1} = 8 } \\ \\ \sf{ Second \ Term \ - } \\ \\ \sf{ S_{2} = 3 {2}^2 + 5 \times 2= 12 + 10 = 22 } \\ \\ \sf{ Common \ Difference \ - } \\ \\ \sf{ => S_{2} - S_{1} = 22 - 8 = 14 } \\ \\ \sf{ Hence , \ d = 6 } \\ \\ \sf{ t_{n} = t_{1} + (n-1) \times d } \\ \\ \sf{ t_{n} = 8 + (n - 1 ) \times 6 } \\ \\ \sf{ => 6n + 2 } \\ \\ \sf{ Hence , \ t_{n} \ is \ 6n + 2 . }$

Answer 2

Given :  Sₙ   = 3n² + 5n

To find : T₁  ,  d  , Tₙ

Solution:

Sₙ   = 3n² + 5n

=> Sₙ₋₁  = 3(n-1)² + 5(n - 1)

Tₙ = Sₙ  -  Sₙ₋₁

=> Tₙ  = 3n² + 5n  - (3(n-1)² + 5(n - 1))

=> Tₙ  = 3(n² - (n-1)² ) + 5( n - n + 1)

=> Tₙ  =  3(2n - 1)  + 5

=> Tₙ  =  6n - 3 + 5

=> Tₙ  =  6n + 2

6 (1) + 2 =  8

d =  Tₙ₊₁ - Tₙ   =  6(n + 1) + 2 - (6n + 2)  =  6

Tₙ  =  6n + 2  , T₁ =  8  & d  =  6

Another Method :

S₁   = T₁  = 3(1)² + 5(1) =  8

=> T₁  = 8

S₂ =  T₁ + T₂ =  3(2)² + 5(2) =  22

=> 8 +  T₂ =   22

=>  T₂ = 14

d = T₂  - T₁   = 14 - 8  = 6

=> d = 6

a = T₁  = 8

Tₙ = a + (n - 1)d   =  8  +  (n-1)6   =  6n + 2

=> Tₙ = 6n + 2

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