Question

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Answer 1

Solution 13.)

T2 = 2

=> a +d = 2 ------(1)

T4 = 10

=> a + 3d = 10 ------(2)

On subtracting equation 1 from 2, we get

2d = 8

=> d = 4

Now,

a + 4 = 2

=> a = - 2

S10 = 10/2[ 2(-2) +(10 - 1)(4)]

= 5 ( - 4 + 36)

= 5 × 32

= 160


Solution 14.)

a = 1

d = 1

Sn = n/2 [ 2(1) + (n-1)(1)]

= n/2 ( 2 + n - 1)

= n/2 ( n +1)

= n/2 ( 1+ n)

Answer 2

Hey !!

2nd term = 2

a + d = 2 ---------(1)

And,

4th term = 10

a + 3d = 10 ---------(2)

From equation (1) , we get

a + d = 2

a = ( 2 - d ) -------(3)

Putting the value of a in equation (2) , we get

a + 3d = 10

2 - d + 3d = 10

2d = 8

d = 4

Putting the value of d in equation (3) , we get

a = ( 2 - d ) = -2.

Sn = n/2 × ( 2a + ( n -1 ) × d

S10 = 10/2 × ( 2 × -2 + ( 10 - 1 ) × 4 .

S10 = 160

14. First term ( a ) = 1

Common difference ( d ) = 1.

Sn = n/2 × [ 2a + ( n - 1 ) × d

Sn = n/2 × [ 2 × 1 + ( n - 1 ) × 1

Sn = n/2 × ( 2 + n - 1 )

Sn = n /2 ( n + 1 )