Progressions Part 1

Algebra Level 2

$54+51+48+45+ \cdots$

You are given the sum of an arithmetic progression of a finite number of terms, as shown above.

What is the minimum number of terms used to make a total value of 513?

The answer is 18.

3 solutions

Sravanth C.
May 15, 2015

We know that, $S_n=\dfrac{n}{2}[2a+(n-1)d]$

Where, $S_n$ is the sum till the $n^{th}$ term, $n$ is the number of the term, $a$ is the first term and $d$ is the difference.

Here, $S_n=513$ , $n=?$ , $a=54$ , $d=(-3)$

Substituting the values in the equation above, we get, $513=\dfrac{n}{2}[2\times(54)+(n-1)(-3)]$ $513=\dfrac{n}{2}[108+(-3n)+3]$ $1026=n(111-3n)$ $1026=111n-3n^{2}$ $-3n^{2} +111n -1026=0$ $3(-n^{2} +37n -342) =0$

Therefore, we settle with the following equation, $(-n^{2} +37n -342) =0$ , by factorizing it, we get, $(-n^{2} +18n+19n -342) =0$ $(-n(n-18)+19(n -18) =0$ $(n-18)(-n+19)=0$

Hence we get two solutions, they are $(n=\boxed{18})$ and $(n=\boxed{19})$ . From which we find that, $\boxed{18}$ is the suitable answer by trial and error method.

Moderator note:

This is the conventional approach. Nicely done. Bonus question: Because the first term and the common difference are a multiple of 3, can you find another approach to this problem?

In response to Challenge master: @Calvin Lin sir, I'd tried but I'm getting this, it's appearing as if I'd done nothing but written it the other way round! Can you help me???

Here's what I'm getting . . .

We can write the sequence as follows, $54+(54-3)+(54-3-3)+(54-3-3-3) +. . .n(times)=513$ $. . . . . . (1)$

Hence, we are adding $54$ , $n$ times, and subtracting $(-3)+(-6)+(-9). . . .(n-1)\quad times)$

Therefore, using $S_n=\dfrac{n}{2}[2a+(n-1)d]$

Where, $n=(n-1)$ , $a=(-3)$ and $d=(-3)$ , we can get the value we are subtracting,

$S_{(n-1)}=\dfrac{(n-1)}{2}[2(-3)+((n-1)-1)(-3)]$

$S_{(n-1)}=\dfrac{(n-1)}{2}[(-6)+(-3n+6)]$

$S_{(n-1)}=\dfrac{(n-1)}{2}(-3n)$

$S_{(n-1)}=\dfrac{(-3n^{2}+3n)}{2}$ $. . . . . . (2)$

Now substituting $(2)$ in $(1)$ ,

$54n+\dfrac{(-3n^{2}+3n)}{2}=513$

$108n+{(-3n^{2}+3n)}=1026$

$-3n^{2}+111n-1026=0$

Similarly, solving the this equation, we get two solutions, of which $n=\boxed{18}$ is correct.

Sravanth C. - 6 years ago

this is simple, but much faster would be, we know after certain point the sum keeps decreasing when we go into the negative terms and the total, becomes depleted by steps of 3. so if we find the sum till 0. which is 19*54/2 but here coincidentally it is 513, now since we are adding 0 as the 19th term, 18 would be the answer since minimum number of terms (this is what I didnt apply and said 19).

Rajasekhara Madhusudan - 4 years, 11 months ago

I thought the answer is 19. But now i'm all clear.

niloy debnath - 4 years, 12 months ago

Vishwarup Kalaikannan
May 18, 2015

54+51+48+45+.... can be written as 3(18+17+16+15....) which is =513.

513=3(171)

18,17,16 are consecutive integers.

Sum of consecutive integers is=n(n+1)/2 where n is the no of integers.

171=n(n+1)/2

n=18

no of integers (terms)=18

How do u know that the progression end in 1.

To use the formula n(n+1)/2 ,first term must be 1.

genis dude - 3 years, 10 months ago

A Former Brilliant Member
Dec 21, 2016

$d=-3$

$S=$ $\frac{n}{2}$ $[2a_1+(n-1)d]$

$513=$ $\frac{n}{2}$ $[108+(n-1)(-3)]$

$n^2-37n+342=0$

$n=18$ $and$ $n=19$

Since it is asking for the minimum, $n=18$

Good. Really helpful.

Jinoshiya Selvaraj - 3 years, 4 months ago

Progressions Part 1

The answer is 18.

3 solutions

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