# I am having problems related to series and sequences and here are those questions.

• July 12th 2012, 10:53 PM
lord287
I am having problems related to series and sequences and here are those questions.
Q1. The sum of first n terms(n>1) of an A.P. is 153 and c.d. is 2. If the first term is an integer, the number of possible values of n is?

Q2. The minimum number of terms of 1+3+5+7+ adds up to a number exceeding 1357 is?

Q3.The number of terms common to two A.P.s 3,7,11,...,407 and 2,9,16,...,709 is?

Q4. An A.P whose first term is unity and in which the sum of the first half of any number of terms to that of the second half of the same number of terms is in constant ratio, the c.d d is?

Q.5 If x,y,z are real numbers satisfying the expression 25(9x² + y²) + 9z² -15(5xy + yz + 3zx) = 0 then x,y,z are in?
a)AP
b)GP
c)HP
d)none

Here are some questions to which I really need an explained answer. I hope I will get answer soon
• July 12th 2012, 11:40 PM
Prove It
Re: I am having problems related to series and sequences and here are those questions
Quote:

Originally Posted by lord287
Q1. The sum of first n terms(n>1) of an A.P. is 153 and c.d. is 2. If the first term is an integer, the number of possible values of n is?

Q2. The minimum number of terms of 1+3+5+7+ adds up to a number exceeding 1357 is?

Q3.The number of terms common to two A.P.s 3,7,11,...,407 and 2,9,16,...,709 is?

Q4. An A.P whose first term is unity and in which the sum of the first half of any number of terms to that of the second half of the same number of terms is in constant ratio, the c.d d is?

Q.5 If x,y,z are real numbers satisfying the expression 25(9x² + y²) + 9z² -15(5xy + yz + 3zx) = 0 then x,y,z are in?
a)AP
b)GP
c)HP
d)none

Here are some questions to which I really need an explained answer. I hope I will get answer soon

What have you tried?
• July 12th 2012, 11:41 PM
richard1234
Re: I am having problems related to series and sequences and here are those questions
Q1: You have $a + (a+2) + \dots + (a + 2n-2) = 153$. There are n terms, so $a$ appears n times. Also, $2 + 4 + \dots + (2n - 2) = 2(1 + 2 + \dots + (n-1) = n(n-1)$. So the LHS is equivalent to

$an + n(n-1) = 153 \Rightarrow n(a + n-1) = 153$. Positive factors of 153 are 1,3,9,17,51,153. However n cannot equal 1, but n can equal any other factor of 153, and there would be an integer value for $a$. So there are five values.

Q2: $1 + 3 + 5 + \dots + 2k-1 = k^2$. What is the smallest k such that $k^2 > 1357$?

Q3: Solve the congruence $n \equiv 3 (\mod 4)$ and $n \equiv 2 (\mod 7)$. By the Chinese remainder theorem, all solutions are congruent mod 77.

Q4: I don't really understand that question, but what if $d = 0$?

Q5: Expand everything out and multiply by 2:
$450x^2 + 50y^2 + 18z^2 - 150xy - 30yz - 90zx = 0$. Note that this factors to

$(15x - 5y)^2 + (15x - 3z)^2 + (5y - 3z)^2 = 0$

It is obvious that $15x = 5y$, $15x = 3z$, $5y = 3z$, because each of the squares must be zero. Solving for the variables yields $x = x$, $y = 3x$, $z = 5x$, so they must be in an arithmetic progression.
• July 13th 2012, 12:19 AM
lord287
Re: I am having problems related to series and sequences and here are those questions
Quote:

Originally Posted by richard1234
Q3: Solve the congruence $n \equiv 3 (\mod 4)$ and $n \equiv 2 (\mod 7)$. By the Chinese remainder theorem, all solutions are congruent mod 77.

Q4: I don't really understand that question, but what if $d = 0$?

THANX SO MUCH

And if u want I can give u the answer to the 4th question the answer to 4th is d=2, If u can do it now plz help me.

And can u explain Q3 with a different method I have not done the Chinese remainder theorem. I am really not understanding what u did in Q3. And given answer of Q3 is 14. Plz give a little more details for these two answers if u can.

Also plz tell why u took out the positive factors of 153 in question 5 and will the integer value of 'a' be fixed or the 5 values of 'n' will be satisfied for any value of 'a'?
• July 13th 2012, 08:15 AM
richard1234
Re: I am having problems related to series and sequences and here are those questions
Q1: $n(a + n - 1) = 153$. Both numbers are (positive) integers so you want to find factors of 153 that satisfy n (greater than 1). Once you fix a factor for n, there is an integer value for $a$.

Q3: You know mods, right? If not, I suggest you look them up.

You can solve the congruence, or you can simply find the first integer that is both 3 mod 4 and 2 mod 7. This happens to be 23. By the Chinese remainder theorem (see link below), all other solutions must be congruent mod 28 (Idk what I was thinking when I said 77...sorry about that). Therefore the solutions are 23, 51, 79, 107, ..., 387 (415 is too large). How many numbers are in the set {23, 51, 79, ..., 387}?

Chinese remainder theorem - Wikipedia, the free encyclopedia

$Q4$: So if you have 2n terms with common difference d, you're saying that $k(1 + (1+d) + \dots + (1+(n-1)d)) = 1 + nd + 1 + (n+1)d + \dots + (1 + (2n-1)d)$?

If so, I can see how d = 2 (since the sum 1 + 3 + ... + 2n - 1 is a perfect square; the first half of the terms sum up to ~1/4 of the total sum). But what about d = 0? 1,1,1,1... is technically considered an arithmetic sequence.
• July 13th 2012, 09:47 AM
lord287
Re: I am having problems related to series and sequences and here are those questions
More questions:-

Q. If a + b + c = 3, a>0, b>0, c>0, then the greatest value of a²b³c² is

Q. The sum of 0.2 + 0.004 + 0.00006 +...... ∞: is

Ans. 2000/9801

Q. The sum pf all the numbers of the form n3 which lie between and 10000 is?

Ans. 53261

Q. If harmonic progression Hn = 1 + 1/2 +.......+ 1/n then value of 1 + 3/2 + 5/3 +.....+ 2n-1/n is
a) Hn + n
b) 2n- Hn
c) (n-1) + Hn
d) Hn + 2n

Sorry if u people are finding these questions by me easy but they r not easy for me since I am only in grade 10+1
• July 13th 2012, 10:10 PM
Prove It
Re: I am having problems related to series and sequences and here are those questions
Quote:

Originally Posted by lord287
More questions:-

Q. If a + b + c = 3, a>0, b>0, c>0, then the greatest value of a²b³c² is

Q. The sum of 0.2 + 0.004 + 0.00006 +...... ∞: is

Ans. 2000/9801

Q. The sum pf all the numbers of the form n3 which lie between and 10000 is?

Ans. 53261

Q. If harmonic progression Hn = 1 + 1/2 +.......+ 1/n then value of 1 + 3/2 + 5/3 +.....+ 2n-1/n is
a) Hn + n
b) 2n- Hn
c) (n-1) + Hn
d) Hn + 2n

Sorry if u people are finding these questions by me easy but they r not easy for me since I am only in grade 10+1

Q2.

\displaystyle \begin{align*} 0.2 + 0.004 + 0.00006 + \dots &= \frac{2}{10} + \frac{4}{1000} + \frac{6}{100000} + \dots \\ &= \frac{2}{10}\left(\frac{1}{1} + \frac{2}{100} + \frac{3}{10000} + \dots \right) \\ &= \frac{1}{5}\sum_{n = 0}^{\infty}\frac{n + 1}{100^n} \\ &= \frac{1}{5}\left[ \sum_{n = 0}^{\infty}\frac{n}{100^n} + \sum_{n = 0}^{\infty}\left(\frac{1}{100}\right)^n \right] \end{align*}

The second sum is geometric. I'm not sure how to evaluate the first...
• July 13th 2012, 10:22 PM
richard1234
Re: I am having problems related to series and sequences and here are those questions
Q3 should be pretty simple. What are all the cubes from 1 to 10000, and how do you sum the first n cubes?

I'm still figuring out Q1 but it probably involves AM-GM inequality.
• July 13th 2012, 10:29 PM
richard1234
Re: I am having problems related to series and sequences and here are those questions
Q2: Series obviously converges. Also, note that

$\frac{2}{10} + \frac{4}{1000} + \frac{6}{100000} + \dots = (\frac{2}{10} + \frac{2}{1000} + \dots) + (\frac{2}{1000} + \frac{2}{100000} + \dots) + \dots$

Factor stuff out.

$= (\frac{2}{10} + \frac{2}{1000} + \dots)(1 + \frac{1}{100} + \frac{1}{10000} + \dots)$

Both series are geometric, it shouldn't be hard to compute the sum.
• July 14th 2012, 01:30 AM
lord287
Re: I am having problems related to series and sequences and here are those questions
Quote:

Originally Posted by richard1234
Q2: Series obviously converges. Also, note that

$\frac{2}{10} + \frac{4}{1000} + \frac{6}{100000} + \dots = (\frac{2}{10} + \frac{2}{1000} + \dots) + (\frac{2}{1000} + \frac{2}{100000} + \dots) + \dots$

Factor stuff out.

$= (\frac{2}{10} + \frac{2}{1000} + \dots)(1 + \frac{1}{100} + \frac{1}{10000} + \dots)$

Both series are geometric, it shouldn't be hard to compute the sum.

Didn't understand how did u factor that stuff out. I had been trying to factor it but could not and I am not understanding how come the way u have written it is the factor of that equation??
• July 14th 2012, 01:31 AM
lord287
Re: I am having problems related to series and sequences and here are those questions
Quote:

Originally Posted by richard1234
Q3 should be pretty simple. What are all the cubes from 1 to 10000, and how do you sum the first n cubes?

I'm still figuring out Q1 but it probably involves AM-GM inequality.

Oh I am sorry I wrote wrongly the real question was sum of cubes between 100 and 10000, not 1 to 10000