Question

kindly answer. !!!!!!​

Answer 1

Solution:-

Given :-

=> T₄ = 8

=> T₆ = 14

To find :-

=> First term ( a )

=> Common Difference ( d )

=> S₂₀

Formula:-

=> Tₙ = a + ( n - 1 )d

Now Take

=> T₄ = 8

=> 8 = a + ( 4 - 1 )d

=> a + 3d = 8 .....( i )eq

Now take

=> T₆ = 14

=> 14 = a + ( 6 - 1 )d

=> a + 5d = 14 .......( ii )eq

Subtract ( ii ) from ( i )

=> a + 3d - a - 5d = 8 - 14

=> - 2d = - 6

=> d = 6/2

=> d = 3

Now put d = 3 on ( ii ) eq

=> a + 5d = 14

=> a + 5 × 3 = 14

=> a + 15 = 14

=> a = 14 - 15

=> a = - 1

Now we have to find S₂₀

Given

=> first term ( a ) = - 1

=> Common difference ( d ) = 3

=> number of terms ( n ) = 20

Formula

=> Sₙ = n/2{ 2a + ( n - 1 )d }

=> S₂₀ = 20/2 { 2 × - 1 + ( 20 - 1 )×3 }

=> S₂₀ = 10 { - 2 + 19 × 3 }

=> S₂₀ = 10 { - 2 + 57 }

=> S₂₀ = 10 × 55

=> S₂₀ = 550

Answer

=> First term ( a ) = -1 , common difference ( d ) = 3 and S₂₀= 550

Answer 2

Answer:

1. first term= -1

2 Common d=3

3. sum of first 20 terms = 550