Answers

2016-02-29T07:24:16+05:30
A=1
d=3
an=X
an=a+(n-1)d
x=1+3n-3
=3n-2
Sn=287
Sn=n/2 (a+an)
=n/2 (1+3n-2)
287 =n/2 (3n-1)
574=3n^2-n
3n^2-n-574=0
now by finding the value of n by quadratic formula . Then substitute the value of n in an
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2016-02-29T13:18:43+05:30
Sn = n/2(2a+(n-1)d)  given a=1, d=4-1=3 & Sn = 287
287 = n/2 (2*1 +(n-1) 3)  
287*2 = n(2 + 3n - 3)
574 = 2n + 3n^2 - 3n
3n^2 -n - 574 = 0
on solving the quadratic equation using formula n= -b + sq.root(b^2 -4ac)
                                                                            -----------------------
                                                                                          2a
we get   n = 14,  -41/3  n not equal to  -41/3 due to negative nos.
n=14
Sn = n/2 (a +l)
287 = 14/2(1 +x)
574 = 14 (1+x)
574 / 14 = 1+x
41 = 1 + x
So, x = 41 - 1
      x = 40 is the solution
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