Question

help me plss my frnds​

Answer 1

Answer

2........

given 3rd term = 4

i.e, a+2d=4 ---- 1

and 9th term = -8

i.e., a+8d=-8 ------------- 2

by by subtacting 1 and 2 we get,

      a+2d = 4

      a+8d = -8

     -    -      +

     ----------------

        -6d = 12

     ---------------

so, d = -12/6

      d = -2

substitute it in (1) we get, 

a+2(-2) = 4

a = 4+4 =8

a=8

here we get a=8

2nd term = a+d = 8+(-2) = 8-2 = 6

3d term = a+2d = 8+2(-2) = 8-4 = 4

4th term = a+3d = 8+2(-3) = 8-6 = 2

5th term = a+4d = 8+2(-4) = 8-8 =0

6th term = a+5d =8+5(-2) = 8-10 = -2 .......

so the series we got is 8,6,4,2,0,-2.....

here we get zero at 5th place

so, 5th number in the series is 0

3.........

Given, a3 = 12 and a50 = 106

a3 = a + 2d = 12

a50 = a + 49d = 106

Subtracting 3rd term from 50th term, we get;

a + 49d – a – 2d = 106 – 12

Or, 47d = 94

Or, d = 2

Substituting the value of d in 12th term, we get;

a + 2 x 2 = 12

Or, a + 4 = 12

Or, a = 8

Now, 29th term can be calculated as follows:

a29 = a + 28d

= 8 + 28 x 2

= 8 + 56 = 64

5.........

Here it is an AP

a = 4

d = 2

n = 10

Sum of the first 10 terms

= (n/2)[2a+(n-1)d]

=(10/2)[(2x4)+(10-1)2]

=5[8+18]

=130

6.......

n=20 , an=57, a=3

sn=n/2 ( a+l)

=20/2(3+57)

=10(60)

=600

Answer 2

which number you want .