At least how many terms in the GP 20+28+36+.... is greater than 1000 ?

#a_0 = 20, r = 8#

#a_n = a_0 + (n-1)r = 12 + 8n#

#S_n = (a_0 + a_n) n /2 = (32 + 8n)n/2 > 1000#

#4n^2 + 16n - 1000 > 0#

#Delta = 16256#

#n_0 = (-16 pm 127.5)/8#

#n_1 = -17.9#

#n_2 = 13.9#

#n < n_1 or n > n_2#

#S_{14} = 1008#

Observe that the given series is Arithmetic with #a=20, d=8#.

Suppose that, #S_n gt 1000#.

Since, #S_n=n/2{2a+(n-1)d]#, we have,

# n/2[2(20)+8(n-1)] gt 1000#.

#:. n/2(8n+32) gt 1000#.

#:. n(4n+16) gt 1000#.

#:. 4n(n+4) gt 1000#.

#:. n(n+4) gt 250#.

#:. n^2+4n gt 250#.

To complete square, we add #4# and get,

# n^2+4n+4 gt 254#.

#:. (n+2)^2 gt 254#.

#:. n+2 gt sqrt254~~15.94#.

#:. n+2 ge 16#.

#:. n ge 14#.

#:. n_min=14#.

Here ,

#S_n=20+28+36+...#

Let , #" first term " a_1=20 and #

#" for common ratio" :r=28/20=7/5 and r=36/28=9/7!=7/5#

So , the given terms are not in #GP#

Let us try for #AP :#

#"Common difference " d=28-20=8 and d=36-28=8#

So, the given terms may be in # AP#

#:.# The sum of first n-term:

#S_n=n/2[2a_1+(n-1)d]=n/2[40+(n-1)8]#

#:.S_n=20n+n(n-1)4]=20n+4n^2-4n#

#:.S_n=4n^2+16n=4(n^2+4n)#

#"For " S_n > 1000# we have

#4(n^2+4n) > 1000#

#:.color(red)(n^2+4n >250#

#n=13=>n^2+4n=13^2+4(13)=color(red)(221#

#n=14=>n^2+4n=14^2+4(14)=color(red)(252#

So ,

#n=13=>S_n=884 < 1000#

#n=14=>S_n=1008 > 1000#

Hence ,there are infinitely many terms for which

#S_n > 1000,where ,n >13#

But we can say that ,

" The least value of #n# for which #S_n > 1000# is #n=14# "