Answers

2015-05-08T16:20:11+05:30
i) given a=7, a13=35
a13 (or)  a+12d = 35 ------- 1
a = 7 (or) a+0d = 7  ------- 2
by subtracting 1 with 2 we get
           
                   a+12d = 35
                   a+  0d = 07
                 -       -      -
                  ------------------
                        12d = 28
                  -------------------
 d = 28/12 = 2.33
so,  d = 2 (approximately)

S13 = n/2 [a+l]
n=13, a= 7, a13 = l = 35
S13 = 13/2 [7+35]
S13 = 13/2 [42]
S13 = 13 [21]
therefore, S13 = 273

ii) S10 = 125, a3 = 15 is given
   S10 = 125 = 10/2 (2a+9d)     [since  a+l means a+a10 = a+a+9d]
   2a+9d=25 ------- 1
   a3 = a+2d = 15 ------ 2
   subtracting 2 from 1 we get, 
                2a+9d-(a+2d) = 25-15
                a+7d=10 
               i.e., a8 = 10   and given a3=15

subtracting a3 from a8 we get
    (a+7d)-(a+2d) = 10-15
    5d=-5
    d = -1
by keeping it in 2 we get
  a+2(-1) = 15
  a= 15+2
  a= 17
 then,
 a10 = a+9d = 17+9(-1) = 17-9 = 8
 therefore, 
 a10 = 8
   

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