Question

please Answer correctly​

Answer 1

Here is your ans.


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Answer 2

Answer:

Step-by-step explanation:

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

NOW,   20/2 [2a + (20-1)] = 400

 10[ 2a + 19d ]  = 400

     2a + 19d  =  40         ---------   (i)

similarly,   For 40 terms 

     40/2[ 2a+39d]  = 1600

       2a + 39d = 80           --------- (ii)

now on subtracting  (i) & (ii)

we get,    20d = 40          

                d =   2

 and therefore,    2a+ 19*2 = 40

                              a = 1

hence sum of 1st ten terms is

       10/2[2*1 + (10-1)2]

        5[ 2 + 18 ] = 20*5

          =  100