Question

Please help me with this question ! ^^"

Answer 1

Hi !

sum of first 6 terms = S₆ = 0.

fourth term = a₄ = 2

i.e ,
a + 3d = 2       ---> [1]
-----------------------

n = 6
S₆ = 0
S₆ = n/2 [ 2a + [ n - 1]d ]
 0    = 6/2 [ 2a + ( 6 -1)d]
 0 = 3 [ 2a + 5d]
0 = 6a + 15d

6a + 15d = 0              ----> [2]

solving equations 1 and 2 :-


6 x (a + 3d = 2  ) => 6a + 18d = 12
6a + 15d = 0 

- 6a + 18d = 12
 6a + 15d = 0 
--------------------
3d = 12

d = 4

a + 3d = 2

a + 12 = 2

a = -10

-----------------------------------------------------------------------

sum of first 30 terms :-

sn = n/2 [ 2a + [n-1]d]
    = 30/2 [ 2*-10 + 29 x 4 ]
    = 15 [ -20 + 116]
    = 15 [ 96]
    = 1440

sum of first 30 terms is 1440

Answer 2

Let the terms be a, a+d, a+2d, a+3d are the first four terms.

So a+3d = 2... (1)

we know that sum of first n terms = n/2[ 2a+(n-1) d ]

0= 3( 2a +6-1d)

2a+5d=0....(2)

(1)✖ 2- (2)

2a+6d = 4

2a+5d=0
============
d =4
=============

Now find the a,

a+3(4)= 2

a= 2-12

a=-10

__________

sum of first 30 terms = 30/2[ 2(-10)+ (30-1)4]

=15( -20+116)

=15( 96)

= 1440

hope helped!!!