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2015-05-10T19:23:46+05:30
Given an = 9-5n
to know the series, 
a1 = 9-5(1) = 4
a2 = 9-5(2) = 9-10 = -1
a3 = 9-5(3) = 9-15 = -6
a4 = 9-5(4) = 9-20 = -11 ......
So the series is 4, -1, -6, -11 ... 
common difference d = a2 - a1 = -1-4 = -5
So, the common difference d = -5      (we can check it by doing a3-a2, a4-a3 ... )
when we do with the a3-a2 and a4-a3 too, we get the common difference d=-5
So the {an} i.e, the series is in A.P
Then, by using
Sn = n/2 [2a +(n-1)d]
In above Calculation, We got that a = 4, d = -5, n=15 (as per problem)
*he wasnt mentioned that in question, he mentioned that in comments
S15 = 15/2 [ 2(4)+(15-1)(-5)]
S15 = 15/2 [ 8+14(-5)]
S15 = 15/2 (-70+8)
S15 = 15/2 (-62)
S15 = 15(-31)
S15 = -465
therefore the sum of the first 15 numbers on the series that we find out is -465
3 5 3
huhhh... it is the answer mr. rajnani.. plzz click on thanks and mark it as best.. hope it helps u...
is the answer was crct??
ofcourse its correct
wait one more que?
ok post that
2015-05-10T19:30:28+05:30
Given an=9-5n
so a1=9-5(1)
        =4
a2=9-5(2)
    =9-10
=  -1
a3=9-5(3)
    = -6

here common difference (d)=a2-a1= -1 -4=-5 
and a3-a2= -6+1
                 = -5
here a2-a1= -5 and a3-a2=-5
as they have the same common difference, they form a.p.

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