Zeeshan A. GCSE Maths tutor, A Level Physics tutor, A Level Maths tut...

Zeeshan A.

£22 - £24 /hr

Currently unavailable: for regular students

Studying: Theoretical Physics with Applied Maths (Masters) - Birmingham University

4.9
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18 reviews| 65 completed tutorials

Contact Zeeshan

About me

Hi,

I myself am studying Theoretical Physics with Applied Maths in the University of Birmingham, and I would absolutely love tutoring (especially Maths!).

I am very patient and motivated - since I know that understanding some concepts can be daunting at GCSE  and A-level.

I aim to understand the student's position whilst approaching concepts in a logical manner at an appropriate pace. Provided I am informed in advance of the topic for an arranged lesson, I will give my time to prepare fully around it before the sessions. And/Or if there are any questions or past-paper help I will guide through the important facts through the paper.

In any case, more importantly, I believe the tutee him/herself should lead the sessions, if they are going to be effective.

Normally in the sessions I would give explanations along with worked examples discussing with the tutee.

In the midst I would ask reasonably progressive questions to make sure the tutee can understand the relevant concepts. 

 I also achieved Silver Award in the Maths Challenge in my year 12.

Availability: Available most evenings except Monday evenings

Hi,

I myself am studying Theoretical Physics with Applied Maths in the University of Birmingham, and I would absolutely love tutoring (especially Maths!).

I am very patient and motivated - since I know that understanding some concepts can be daunting at GCSE  and A-level.

I aim to understand the student's position whilst approaching concepts in a logical manner at an appropriate pace. Provided I am informed in advance of the topic for an arranged lesson, I will give my time to prepare fully around it before the sessions. And/Or if there are any questions or past-paper help I will guide through the important facts through the paper.

In any case, more importantly, I believe the tutee him/herself should lead the sessions, if they are going to be effective.

Normally in the sessions I would give explanations along with worked examples discussing with the tutee.

In the midst I would ask reasonably progressive questions to make sure the tutee can understand the relevant concepts. 

 I also achieved Silver Award in the Maths Challenge in my year 12.

Availability: Available most evenings except Monday evenings

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Ratings & Reviews

4.9from 18 customer reviews
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Luke (Student)

October 26 2015

Very helpful and patient, Thanks

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Sheeru (Student)

April 26 2015

He is excellent. Very helpful.

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Iftikhar (Parent)

March 31 2015

Really an amazing tutor. Has helped me so much! Very dependable, patient and clearly explains all questions and answers.

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Sheeru (Student)

April 4 2015

He is excellent. Very helpful.

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathsA-level (A2)A
PhysicsA-level (A2)A

General Availability

Before 12pm12pm - 5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
Further Mathematics A Level£24 /hr
MathsA Level£24 /hr
PhysicsA Level£24 /hr
MathsGCSE£22 /hr

Questions Zeeshan has answered

How to derive the formula for a geometric series sum

Any geometric series is defined by an initial term a, and a common ratio b

This means that we start with a and multiply by b to get each next term.

And so the general geometric series is written:

a*r^0,a*r^1,a*r^2,a*r^3,...,a*r^n ; where n+1 = no. of terms.

Now the sum of the above, S is:

S = a*r^0+a*r^1+a*r^2+a*r^3+...+ar^n

Factorise r out in most terms in right hand side:

S = a*r^0+r*(a*r^0+a*r^1+a*r^2+...+a*r^(n-1))

Now note that the sum within the brackets includes all terms in our original sum S, save the last term a*r^n. This means we can substitute (S-a*r^n) for it. Therefore,

S = a*r^0+r*(S-a*r^n)

Now all we are left to do is make the value of the sum, S, the subject of the equation:

S = a+r*S-a*r^(n+1)

S-r*S = a-a*r^(n+1)

S*(1-r) = a*(1-r*(n+1)); factorize S on LHS, and a on RHS.

S = a*(1-r*(n+1))/(1-r)

 

Any geometric series is defined by an initial term a, and a common ratio b

This means that we start with a and multiply by b to get each next term.

And so the general geometric series is written:

a*r^0,a*r^1,a*r^2,a*r^3,...,a*r^n ; where n+1 = no. of terms.

Now the sum of the above, S is:

S = a*r^0+a*r^1+a*r^2+a*r^3+...+ar^n

Factorise r out in most terms in right hand side:

S = a*r^0+r*(a*r^0+a*r^1+a*r^2+...+a*r^(n-1))

Now note that the sum within the brackets includes all terms in our original sum S, save the last term a*r^n. This means we can substitute (S-a*r^n) for it. Therefore,

S = a*r^0+r*(S-a*r^n)

Now all we are left to do is make the value of the sum, S, the subject of the equation:

S = a+r*S-a*r^(n+1)

S-r*S = a-a*r^(n+1)

S*(1-r) = a*(1-r*(n+1)); factorize S on LHS, and a on RHS.

S = a*(1-r*(n+1))/(1-r)

 

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