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

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

3 Answers

Fourteen

Explanation:

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

Aug 11, 2018

14.

Explanation:

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.

Aug 11, 2018

Please see below.

Explanation:

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 "