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

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

Aug 11, 2018

Fourteen

#### Explanation:

${a}_{0} = 20 , r = 8$

${a}_{n} = {a}_{0} + \left(n - 1\right) r = 12 + 8 n$

${S}_{n} = \left({a}_{0} + {a}_{n}\right) \frac{n}{2} = \left(32 + 8 n\right) \frac{n}{2} > 1000$

$4 {n}^{2} + 16 n - 1000 > 0$

$\Delta = 16256$

${n}_{0} = \frac{- 16 \pm 127.5}{8}$

${n}_{1} = - 17.9$

${n}_{2} = 13.9$

$n < {n}_{1} \mathmr{and} n > {n}_{2}$

${S}_{14} = 1008$

Aug 11, 2018

$14$.

#### Explanation:

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

Suppose that, ${S}_{n} > 1000$.

Since, ${S}_{n} = \frac{n}{2} \left\{2 a + \left(n - 1\right) d\right]$, we have,

$\frac{n}{2} \left[2 \left(20\right) + 8 \left(n - 1\right)\right] > 1000$.

$\therefore \frac{n}{2} \left(8 n + 32\right) > 1000$.

$\therefore n \left(4 n + 16\right) > 1000$.

$\therefore 4 n \left(n + 4\right) > 1000$.

$\therefore n \left(n + 4\right) > 250$.

$\therefore {n}^{2} + 4 n > 250$.

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

${n}^{2} + 4 n + 4 > 254$.

$\therefore {\left(n + 2\right)}^{2} > 254$.

$\therefore n + 2 > \sqrt{254} \approx 15.94$.

$\therefore n + 2 \ge 16$.

$\therefore n \ge 14$.

$\therefore {n}_{\min} = 14$.

Aug 11, 2018

#### Explanation:

Here ,

${S}_{n} = 20 + 28 + 36 + \ldots$

Let , $\text{ first term } {a}_{1} = 20 \mathmr{and}$

$\text{ for common ratio} : r = \frac{28}{20} = \frac{7}{5} \mathmr{and} r = \frac{36}{28} = \frac{9}{7} \ne \frac{7}{5}$

So , the given terms are not in $G P$

Let us try for $A P :$

$\text{Common difference } d = 28 - 20 = 8 \mathmr{and} d = 36 - 28 = 8$

So, the given terms may be in $A P$

$\therefore$ The sum of first n-term:

${S}_{n} = \frac{n}{2} \left[2 {a}_{1} + \left(n - 1\right) d\right] = \frac{n}{2} \left[40 + \left(n - 1\right) 8\right]$

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

$\therefore {S}_{n} = 4 {n}^{2} + 16 n = 4 \left({n}^{2} + 4 n\right)$

$\text{For } {S}_{n} > 1000$ we have

$4 \left({n}^{2} + 4 n\right) > 1000$

:.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 ,

$n = 13 \implies {S}_{n} = 884 < 1000$

$n = 14 \implies {S}_{n} = 1008 > 1000$

Hence ,there are infinitely many terms for which

${S}_{n} > 1000 , w h e r e , n > 13$

But we can say that ,

" The least value of $n$ for which ${S}_{n} > 1000$ is $n = 14$ "