find the sum of the first 15 terms of an A.P. whose nth term is 9-5n.?

Answer 1

nth term is 9 - 5n
first term: put n = 1: 9-5 = 4
a=4
second term: put n = 2: 9-10 = -1
d = -5

sum of the first 15 terms = (n/2)*(2a+(n-1)d) = (15/2)*(2*4-14*5)
=(15/2)*(-62)
= -465

Answer 2

The first term is 9-5*1=4
The last term is 9-5*15 = -66
Thus Sum = (15/2)(4-66) = (15/2)(-62) = 15*-31= -465

Answer 3

given nth term is 9-5n.
substituting n=1 we get the first term a(4),
n=2,we get t1(-1) ,
the difference (t1-a)(-1-4) gives d(-5),the commondifference. n=15 we get the last term L(-66)
sum of terms=n/2(a+L)
where a=4,n=15,l=-66
substituting,we get
sum of terms=15/2(4-66)
=15/2(-62)
=15*(-31)
=-465

Answer 4

nth term = 9-5n
So, 1st term = 9 - 5.1
= 4
And 2nd term= 9 - 5.2
= 9 -10
= -1
Also 3rd term = 9 - 5.3
= 9 - 15
= -6
So A.P. = 4,-1,-6................
Common Difference = -5
So,sum of first 15th terms of A.P.= n/2[2a+(n-1)d]
= 15/2[2.4+(15-1)-5]
= 15/2[8+(14)-5]
= 15/2(8- 70)
= 15/2.( -62)
= 15. (-31)
= - 465

Answer 5

OK!

The idea here is to look at the big picture:

SUM (9-5n) for n=1 to 15

= [9*15] - [5 * (SUM n, for n=1 to 15)]

= 135 - (1+2+3+...+15)

= 135 -120

= 15

Done.

Answer 6

-465.easy put n=1,2.find first term second and hance common difference -5 use the formula Sn=n/2(2a+(n-1)d)and solve
n=15 a=4 d=-5

Answer 7

nth term- 9-5n
when n=1, 9-5x1=4
when n =2, 9-5x2= -1
when n = 3, 9-5x3= -6
a=4, d= -5 , n= 15
an= a+(n-1)d
4+(14) -5
-66
S15= n/2 {a+an}
15/2{4-66}
= -465