So, above is the question for the resolved response below:
https://1drv.ms/u/s!AvbtopNu2rC6jYoWDAZ4Xmf0pZwpXA?e=5hPtG6 -- Wanna hear?
Read it out!
Take out the periodic table (and any ionization energies – related content); have them handy.
Some things about the scenario, the 1st IE of Mg would be greater than its previous elements since the electron configuration tells it all. Look into the 3s electron shell. It's about staying away from the nucleus more and more that makes IE go high. Do revisit everything about IE once.
This question is more qualitative in the sense that you got graphs that you need to closely observe and relatively analyse.
Do note down the 1st IE of Magnesium first. And hey, it’s 738 kJmol – 1 for Mg.
As you go up the ionization energies, it's valence and then the inner electrons. To bifurcate that way, look for the “large” “jump”. Thankfully, don’t you see a major increase across the ionization energies once you’re going ahead of the 2nd graph and the 3rd graph one goes up? So, the large jump is right after the 2nd ionization energy of element B; do observe that, relatively, it’s only element B. So hey, It’s G2 of 2 valence electrons you need to focus on. And you do know that Mg is out there! Relatively, only element B has such graphs behaving in a way that goes with group 2.
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SECTION – A
1)
An arithmetic sequence (or progression) is a sequence in which every two successive terms have a constant common difference.
All you know is S40 = 1900 and u40 = 106 . All you need to find is u1 So its time to determine the sequence for it, so we need to figure out the general rule.
Adding or summing first forty terms or values in the sequence yielded 1900. The 40th term of the sequence is 106. That said, you got to find the first term, common difference, so that you’d have one variable to solve in the equation for Tn or in other words , so that you can calculate “ n ” .
Note that:
T is the term ,
n is the number of terms in sequence
Tn is nth term
So a sequence like “ T1 , T2 , T3 ,…..,Tn ” works just like “ a1 , a2 , a3 , …. , an” does. It’s just a matter of preference. I would’ve solely gone with “ T1 , T2 , T3 ,…..,Tn ”
Here’s to mention one of the things I’m recalling that since n can only be an integer, so there would times when you need to thoughtfully round-off.
Finding the term at a certain location (i.e. the nth term = notation for the term that could be located anywhere so we got variable here) in the sequence
And it’s vice-versa, etc.
Simply put, it’s one thing to find n and it’s the other thing to find Tn
For example,
What is the 7th term of 2 , 4 , 6 , 8,…. ?
Don’t we have those dots ? It means the sequence can go on and on.
7th term implies n = 7
As we know Tn = a + (n – 1) d, where a (or u , u 1) is the 1st term (i.e. mathematically, n = 1) , d is the common difference and Tn is the term at a certain term, i.e. n ; It’s nothing but the nth term of an Arithmetic sequence , determine the general equation(or formula, rule,…) for a sequence with common difference of 2
Vice – versa ?
Does the aforementioned sequence have the value 12 in it ?
We know the general equation already , so equate it to 12 , in order to find the “ n ” in it , i.e. 2 + (n – 1) 2 = 12
That’s what the vice-versa meant.
You can also be given with no sequence as such (a sequence with some variables to be found) but have values for nth terms like that of a30 , a32 , etc .
Now find a , d and a75
Yes, you don’t need to know a31 but you know the preceding and following terms so think about the crux of the common difference and find d .
(a32 – a31) = (a31 – a30) = d
And you can find a75 or anything else for that matter once you determined the general equation ; thanks to a & d .
What if you know a34 instead of a32 ?
Firstly, since you don’t the none of the 3 terms between them , call them by some names, 3 variables !
First term would be as it is, follow it with, say, “p , q , r” and again the 5th term would be as it is.
There is one common thing that relates these terms , i.e., the common difference d . So, mathematically express that relation with a set of expressions , which finally equate to d . There you go, one semi-long equation ! The thing is difference between the successive terms gives you a set of expressions that you can equate with one another because it is the common difference .
Finding d is no different from what we’ve just done in the previous context (where a34 wasn’t the thing but a32 was).
This time the denominator changes to a bigger number cause it’s 5 terms overall. The numerator would be noting but subtracting the first term from the fifth term .
And what about the denominator ?
Now, having 5 terms overall means you could formulate a set of 4 expressions , i.e. (n – 1) = 5 – 1 = 4
Play with this set of expressions in a way that takes you to close to forming an equation at least, where you can substitute the d you found and where there is q but not p or anything else for that matter, because we want to determine the term q ; isolate it . Always, just make sure you’d have one variable to find at a time and accordingly, isolate.
For example
12 , p , q , r , 63
d = =
Oh yes , d doesn’t have to be an integer all the time ! It’s d we are into , not n
Given the sequence, an arithmetic sequence known for having common difference :
63 – r = r – q = q – p = p – 12 = d =
We’d like to find q and we have d, so let’s choose :
p – 12 = ; p =
So , q – = ; q =
Additionally, we can also explore how sigma notation works. And there can be a ton of supercool word problems so wholesome.
Now, have a look at this :
Let the Arithmetic sequence or AP be :
a, (a+d) , (a+2d), (a+3d), (a+6d) , (a+7d), ….. , upto n terms
Let T1 be the first term and hence, Tn = nth term
We know that T1 = a + 0d, T2 = a + 1d , T3 = a+2d , T4 = a+3d , T5 = a+4d , T6 = a+5d, ……and so on up to nth term
Note that the green digits above = the corresponding red digit – 1
So, Tn = a + (n – 1) d – eq(1)
Therefore, T2 – T1 = T3 – T2 = Tn – T(n – 1) = d
What if you want to know the sum of all the terms until the 30th term , i.e. S30 ? You could get a sequence with some terms to start and end with. By the way, an Sn and even un don’t have be given directly as digit(s) but expressions ! You won’t just determine the general equations and face expression kinda ones with variables and all……What if you have expressions but not any digit(s) for the first term , sum of n terms (so n could be anything for that matter) ? So, find d and un of any n ; not impossible !
For now, note that it’s better called the arithmetic series as soon as you’re into figuring out the Sum.
Sn = [ a + an ] – eq(2)
And if you think about it, things make sense.
You begin with the first term a and go up until the last term an , so by adding them , you now multiply it with the n number of terms (cause you don’t know how many they could be and that’s how a variable is sensible here) ,…. Thinking about where’s the 2 from ? You’d know that ahead.
Its absolutely recommended that the denominator of 2 only stays under n ; Above equation is just another form for a well-known form coming ahead . 2 definitely doesn’t come from any average, arithmetic mean of a and an just because these both terms and that 2 are in
an expression .
The takeaway is to let it be
Consider this :
1 , 5 , 9 , 13 , ….. , 77
You got your paper, pencil and you need to find the sum without adding them like you usually do. You can figure out an approach, i.e., an equation that models it, which is what exactly the above equation is, a general equation itself that you’d want to have.
Once again….Note that you are interested in finding the sum of all the n terms in there.
By the way, can you observe that eq(2) can be expressed in another form?
Let’s actually learn more about it and doing so includes the story of eq(1) too
Write down the sequence of terms ranging from the first term a to the last term an . A couple of terms before the “ +… +” begins and after it begins.
The term just before the last one is a( n – 1 ) ;
Now, write the above sequence in the form that incorporates the common difference. This would be eq(3) . So, as you know what d really is, begin with writing the first term as usual and produce a sequence for which you need to answer “doing what to the first term for every next term yields each term of the sequence ?”. That way, you could write a few terms as we start off until you can have “ +….+ ” and write the last terms to end with. And wait, what do you think of the last term ? Is it familiar?
Isn’t it nothing but the nth term of an arithmetic sequence ? And what would the term before this be ?
Let eq(3) pop up here. Before doing so, why don’t you choose a notably wide page or write horizontally? Better have it spacious.
Give out a sequence with the last term in place of the the first and the first term being the last. What about all the terms in between ?
A quick recap ; The thing is, when incorporating d into the scene , you’re expanding each of the terms and technically writing out the RHS of those like a( n – 1 ) , etc , as you are aware of the general equation for the nth term of a sequence. The an would be as it is in every term but what does change is the sign before the coefficient of d in each of the terms . It sure makes sense as you bring out the very fundamental thing about common difference d .
We’re good with eq(3) & eq(4) now.
By the way, you can have as many as terms as you want between “ +…..+ ” . Have more to gain better clarity on the pattern and feeling ease in writing out the terms. Once you’re good with what you’re seeing, let the equations have same number of terms to add the corresponding terms without any mess !
So yes, add the aforementioned equations with each other. Simplify, if any. According to what we’re into finding , let that alone be the LHS.
Doing so changes the LHS from Sn to ……..(you know that). And as far as the RHS is concerned, you should be ending up with the same term repeating over and over again, i.e. (a+an) , after you went on adding the respective terms, the equations with one another. Imagine you have n number of (n+1) terms ? Straight-off multiply them. And that 2 now gets its due ! Sn became all alone ….
When you have something like Sn = [ 2a + (n – 1)d ] and you want to incorporate Tn here , instead of trying to remove that 2 of 2a , with multiplying, taking common, LCM , etc , what’s the best way in which you can incorporate that ?
Umm, could you write 2a in such a way that it works well ?
Always remember to simplify whenever you can unless otherwise mentioned.
And, the words like after, before , above and below could matter so revisit the question and given data!
One of the good habits in these situations is to equate them both so that you can verify if you went right or not !
Also, what if you got some equation, working out which gives another form ? You find it a hassle in working with the equation to get that form, so just do the opposite and then have a look at it. Finally, “what if you got some equation, working out which gives another form?” is sought out.
“Oh well, so much has happened”. Neither did I think of this much being written off 😊 ??
Hope everything so far could ignite refreshing thought processes and unlock the activation energy to explore the questions themselves…! Good with what all has been there so far ? Pat yourself on the back.
Now, marching towards the question….
Oh we weren’t given with a simple, cute sequence . Though, we know the term of n when it is 40 and the sum of terms until 40 .
What is the first term (u1) & common difference (d) ?
First, find u1 and then, d
See, you got an equation as follows:
I’d say, expand it
First, deal with S40 = 1900
How?
What exactly is Sn ? Expand it too. What is it when n = 40 ? Plug in this number into what you’ve expanded , i.e. the equation for Sn
And finally equate the equation for S40 to the term given , i.e. 1900
Likewise, deal with u40 = 106 now :
Expand Tn when n = 40
Equate that to 406
We got 2 equations now, very naturally and in a more relatable fashion. And that’s fine.
But this ain’t it.
If you think about the equation you got for S40 , a feasible form to write it in is eq(2)
We were exactly given the value for un , so you’d only be left with finding u1
Now go back to the 2nd equation, the one you got for u40 . Write it down. You should only have d as the one to be determined now.
2)
From the given data:
The first term, u1 = – 7
(I’d call it u1 , accordingly . Though it can be those quite popular notations like a and T1)
S20 = 620
(a) Again elaborate, expand it’s LHS to try to make sense of it . As we need to find d , after doing so, you know now !
(b) Determine u78 , i.e. the 78th term of the sequence :
Get going with the un equation
What is n here ?
Straight off substitution.
3)
(a) Recall Binomial theorem.
By the way, by any chance if any way like this is popping-up in mind , keep it aside. Don’t do it like (x – 2)2 (x – 2)2 , forget about elaborating it like that even more. Go for binomial theorem.
Note that = ,
According to the given context, it’s
Therefore,
(x – 2)4
=
= 1x4 + 4x3 ( – 2 ) + 24x2 + 4x ( – 2)3 + ( – 2)4
Now , go on ; simplify.
b) Multiplying them off of one another and then, finding where x3 is – Nah
We know that
(3)
(a) Recall Binomial theorem.
By the way, by any chance if any way like this is popping-up in mind , keep it aside. Don’t do it like (x – 2)2 (x – 2)2 , forget about elaborating it like that even more. Go for binomial theorem.
Note that = ,
According to the given context, it’s
Therefore,
(x – 2)4
=
= 1x4 + 4x3 ( – 2 ) + 24x2 + 4x ( – 2)3 + ( – 2)4
Now , go on ; simplify.
(b) Multiplying them off of one another and then, finding where x3 is .
Keep this in mind:
(4)
(a) r ?
Time for division. It’s the common ratio , right?
(b)
T10 ?
We know that Tn = a(r)n – 1
Both n and a are given ; direct substitution
(c)
Recall S = , - 1 ≤ r ≤ 1 ?
(5)
We know that when it’s about the Independent events , this comes up:
P(A B) = P(A) P(B)
Note that should take you to “ or ” , + ; And …..?
We know that P(B) = 2P(A) , so P(A B) could be formed with probability of an event to be found.
Now, what is P(A) ?
Think about this:
P(A B) = P(A) + P(B) – P(A B) ; 0.52 = P(A) + 2P(A) – P(A) [ 2 P(A) ]
Go on !
Finally, what is P(B) ?
(6)
(a)
P ( A is the intersection of 2 events, one of which is the complementary event
Firstly, what exactly is the event A ?
A = A (B = ( A B ) ( A
So now we are well aware of the equation for P(A)
And P ( A has to be on the way
(b)
Following a) and = P( , calculate P (A I B') =
Overall, draw Venn diagrams to visualize better
(7)
(a)
It is their intersected region that counts .
got to take you to “ × ”
(b)
Here, its “ + ” instead
(c)
(i) & (ii)
So, what area is the intersected region of A & (not B) ?
What is not B happens to be everything but it. Out of the whole Venn diagram, wherever it is B , we aren’t up for it.
Just as you see & , you understand. So , what is another form of writing P ( A ?
Here its “ × ”
Do note that neither P(A) nor P(B) can go beyond 1
(8)
(a)
We know that = 1
Sum all those in the 2nd row and equate it to 1
Oh you know k now !
(b)
We know that E(X) =
Multiply the corresponding values with one another and plug in the k found in a)
(9)
You see the exponent is 8 so you know what’s n
Go for
Until n – r = 0 ; r = 8
What would you do to get the constant ? What’s r ? For that, you’d need to figure out when would you get a constant , i.e. x 0
x – 2 , n = 7 ; That is it. The thing is, lay them out to verify when would x 0 be achieved.
Thanks to the 7th term, simplify it well and then, the coefficient of x – 2 equated to 16128 closes the matter !
(10)
(a)
(i)
Come on , the common ratio …
(ii)
With the r found , it’s time for n = 5
(b)
(i)
We need the first term , so what should n be ?
(ii)
Gotta find k , so isolate it
Time for ln
Thanks to (b)(i) and v2 , k could be determined
(c)
What have been found so far, including the value of k , would be computed in the place of the un & vn
Also, do notice the change of the sign when solving that out if any.
And rounding off is important
P.S.
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