Answers

2016-03-25T16:31:40+05:30
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
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  • qais
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2016-03-25T16:42:19+05:30
Let the first term be a and common difference is d
A/q
S20 = (20/2)[2a+(20-1)d]= 400
⇒2a +19d = 40_____(1)
also, S40 =(40/2)[2a+(40-1)d]= 1600
⇒2a +39d = 80____(2)
subtracting (1) and (2), we get
d =2
so, a= 1
S10 = (10/2)[2×1 + (10-1)2] = 5(2+18) = 5×20 = 100

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