Question

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 ?`

Answer 1

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`

Answer 2

`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`.

Answer 3

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` "