# Pls Help Solve These Math Questions

Hello Brilliantians

I do not know how to solve the following questions. I request the readers to kindly help me with solutions and the method of solving them.

$1. \quad The \quad 91st \quad term \quad of \quad the \quad sequence \quad 3, \quad 7, \quad 14, \quad 24, ... \quad is$

$2. \quad The \quad radius \quad of \quad a \quad circle \\ \quad is \quad 25cm. \quad The \quad radii \quad of \quad 3 \quad concentric \quad circles \quad drawn \quad in \quad such \quad \\ a \quad manner \quad that \quad the\quad whole \quad area \quad is \quad divided \quad into\\ \quad 4 \quad equal\quad parts \quad are:$

(a) 25, 25, 25

(b) 25, 50, 75

(c) 25, 30, 35

(d) 25. 25$\sqrt2$, 25$\sqrt3$

$3. \quad Number \quad of \quad real \quad roots \quad of \quad {x}^{2}+3|x|+2=0 \quad is$

$4. \quad If \quad one \quad root \quad a{x}^{2}+bx+c=0 is \quad the \quad square \quad of \quad the \quad other, \quad then$

(a) ${a}^{3}+{b}^{3}+{c}^{3}=3abc$

(b) ${a}^{2}+{b}^{2}+{c}^{2}=3abc$

(c) ${a}^{2}c+{b}^{3}+a{c}^{2}=3abc$

(d) ${a}^{2}c-{b}^{3}+a{c}^{2}=3abc$

$5. \quad If \quad the \quad ratio \quad of \\ \quad sum \quad to \quad 'n' \quad terms \quad of \\ \quad 2 A.P.s \quad (3n+4):(n+3), \quad then \quad the \quad ratio \quad of \quad \\ their \quad 'n'th \quad terms \quad is$

$6. \quad A \quad girl \quad writes \quad all \quad the \\ \quad natural \quad nos. \quad from \quad 100 \quad to \quad 999. \quad The \quad no. \quad \quad of \quad zeroes \quad \\ that \quad she \quad uses \quad is \quad x, \\ \quad the \quad no. \quad of \quad 5's \quad that \quad she \quad uses \quad is \quad y \quad and \quad \\ the \quad no.\quad of \quad 8's \quad that \quad \\ she \quad uses \quad is\quad z. \quad What \quad is \quad the \quad value \quad of \quad 2x+2z-3y$

$7. \quad The \quad remainder \quad when \quad {2}^{89} \quad is \quad divided \quad by \quad 89$

$8. \quad The \quad co-efficient \quad of \quad x \quad in \quad the \quad expansion \quad of \quad \\ (1+x)(1+2x)(1+3x)....(1+100x) \quad is$

Ritu Roy Note by Ritu Roy
5 years ago

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Method to solve first question :

1. However, there are many possibles sequences of such terms. But I solve it by assuming it to be second order A.P. which means second ordered difference is constant in the second order A.P.

i.e. $\quad \quad \quad \quad \quad 3, \quad \quad 7, \quad \quad 14, \quad \quad 24, ...........$

Ist order difference. $\quad 4,\quad \quad 7, \quad \quad 10, \quad \quad 13,.........$

2nd order difference $\quad \quad 3, \quad \quad 3, \quad \quad 3, ..............$

So here in this case, second order difference is constant.

if kth order difference is constant, then the general term is a polynomial expression of kth degree.

So, $t(n)$ is the general term of the given sequence, so $t(n)=an^2+bn+c$.

We know the values $t(1),t(2),t(3)$, from here find the values of $a,b,c$.

$t(n)=\dfrac{3n^2-n+4}{2}$. Now you can find whatever term you want by putting that value of $n$. @Ritu Roy

- 5 years ago

Thanks a lot @Sandeep Bhardwaj .

- 5 years ago

For 3,

$x^2\geq 0; 3|x|\geq 0; 2>0$

Adding them gives $x^2+3|x|+2>0$, so no solution

- 5 years ago

For 8,

Coefficient of $x=1+2+...+100=\dfrac{100×101}{2}=5050$

Explanation: To get power of x as 1, you will have to take "nx" from one term and "1" from other 99. So it would become $x+2x+3x+...+100x$

- 5 years ago

Probably, I can give you the answer of Q.7

According to Fermat's little theorem,

$a^{p-1}\equiv 1\pmod {p}$ where $p$ is a prime.

So, $2^{88}\equiv1\pmod{89}$ since $89$ is a prime .

Or, $2^{89}\equiv2\pmod{89}$

Hence the required remainder is $2$

- 5 years ago

Good! I just think that there is a minor error in mentioning what the theorem is, but I see where you are going with this.

- 5 years ago

$89$ is also a prime number. So, by one of my most favorite mod theorems, Euler's totient function, the expression will just reduce $2^{89} \equiv2^{89-88}\equiv 2 \pmod{89}$

- 5 years ago

It works as well..

- 5 years ago

Shouldn't it be $p-1$? I just realized it's just the same.. lol

- 5 years ago

1. With the assumption that $a_1$ = 3, we can rewrite the sequence of the next term as:$a_{n}$ = $a_{n-1}+(3n-2)$ for $n \geq 2$. I found that with substitution, you can find the nth term of the sequence by $(a_{n}) = (a_1) + 3(1+2+...+(n-1)) + (n-1)$ for $n \geq 1$. So, for example, $a_{2}$ = $a_1$ + $3(2-1)$ + $2-1$ = $3 + 3 + 1 = 7$, $a_{3}$ = $a_1$ + $3(1+2)$ + $3-1$ = $3 + 9 + 2 = 14$, and $a_{4}$ = $a_1$ + $3(1+2+3)$ + $4-1$ = $3 + 18 + 3 = 24$, which are the second, third, and fourth terms of the sequence respectively. So, $a_{91}$ = $a_1$ + $3(1+2+...+89+90)$ + $91-1$ = $3 + 12285 + 90 = 12378$. Therefore, the 91st term of the sequence is 12378. Sorry for my bad typing of the solution. I'm pretty sure others can do better, but I hope this helps. I will probably clarify on how to get to find the nth term of the sequence in a separate note. EDIT: made a major/minor error in the calculation, fixed it now.

- 5 years ago

Nice work! I will just put your idea in a better way. Just analyse the sequence in the following manner. Form a new sequence whose elements are the difference of the adjacent numbers in the given sequence. So the new sequence will read out as $4,7,10 \dots$. It is easy to realise that it is an arithmetic progression. So working backwards, the terms of the given sequence must be related to the sum of terms which form an AP. So for the AP, characterised by the initial term as $1$ and common difference as $3$ let $S(n)$ denote the sum upto $n$ terms. So just by observing it can be realised that the general term of the given sequence is $2 + S(n)$. Using standard formulae evaluate$S(n)$ and hence the required term.

- 5 years ago

Thank you!

- 5 years ago

@Ritu Roy The answer to the 4th question is (c)${ a }^{ 2 }c\quad +\quad { b }^{ 3 }\quad +\quad { c }^{ 2 }a\quad =\quad 3abc$

- 5 years ago

Can you pls elaborate?

- 5 years ago

I guess these are standard board examination questions...Right? Anyway, since no one has answered #5) , I'll post a solution to that one!.

Sum of an A.P upto N terms is :- $\dfrac{n}{2}*( {2a+(n-1)d})$ where the other symbols have their usual meanings. Substitute this with the ratios namely :- $(3n+4):(n+3)$ ..Taking the a's as a1 and a2 and d's as d1 and d2.: we get it as - \dfrac{2a_{1}+(n-1)*d_{1}}{2a_{2}+(n-1)*d_{2}= ratio given. Solve from here using the fact that $a+(n-1)*d$ is the $n$th term of an A.P

- 5 years ago

A better way to say it is that sum of $2m +1$ terms of an AP is $2m +1$ times the $m^{\text{th}}$ term. So the ratio of $n^{\text{th}}$ terms is same as the ratio of their sums upto $2n+1$ terms.

- 5 years ago

W.R.T #2) the question isn't stated clearly as to where the circles are drawn...

- 5 years ago

@Krishna Ar That was how the question was stated.

- 5 years ago

Is the answer of 5th question $\frac{3n-1}{n+1}$

- 5 years ago

Pls help

- 5 years ago

The only help I can offer you is this:

- 5 years ago

Thanks to @Pranjal Jain @Anik Mandal , @Keshav Tiwari , @Sudeep Salgia , @Krishna Ar , @Rajdeep Dhingra , @Marc Vince Casimiro , @Sandeep Bhardwaj and others for your detailed solutions.

I also need the solutions for the ${2}^{nd}$, ${4}^{th}$ and ${7}^{th}$ questions. Thanks a lot once again.

- 5 years ago

For number 4, assume that the roots of that equation were $s$ and $s^{2}$; thus, by Viete's formulas, $b$ = $-(s + s^{2})*a$ and $c$ = $(s)(s^{2})(a)$ = $(s^{3})*(a)$. Then you can substitute into each of the answer choices to get the answer: letter $(c)$

- 5 years ago

@Ritu Roy None of the options in ques no 2 are right . The answer is $\frac { 25 }{ 2 } ,\frac { 25 }{ \sqrt { 2 } } ,\frac { 25\sqrt { 3 } }{ 2 }$

- 5 years ago

Ritu Roy Here is the solution image equal all the area and you will get.the answer. Any problem just reply

- 5 years ago

Thanks a lot.

- 5 years ago

to improve your skill,analyse the questions..... understand the concept and analyse it

- 4 years, 12 months ago