As an experienced tutor registered on UrbanPro, I'd be glad to help with this problem. UrbanPro is indeed an excellent platform for online coaching and tuition, providing students with access to quality education from skilled tutors.

Let's prove the given relation:

Given that the 5th, 8th, and 11th terms of a Geometric Progression (GP) are pp, qq, and ss respectively.

We know the formula for the nth term of a GP is an=ar(n−1)an=ar(n−1), where aa is the first term, and rr is the common ratio.

So, for the 5th term, a5=ar4=pa5=ar4=p
For the 8th term, a8=ar7=qa8=ar7=q
For the 11th term, a11=ar10=sa11=ar10=s

Now, let's form two equations using the given information:

1. From the 5th and 8th terms: q=ar7q=ar7

2. From the 8th and 11th terms: s=ar10s=ar10

Now, let's divide equation 2 by equation 1:

sq=ar10ar7=r10−7=r3qs=ar7ar10=r10−7=r3

So, we get q2=psq2=ps, which is the relation we wanted to prove.

Thus, it's verified that q2=psq2=ps holds true based on the given information. This proof reinforces the properties of geometric progressions and underscores the importance of understanding their terms and ratios. If you have any further questions or need clarification, feel free to ask! And remember, UrbanPro is here to support your academic journey every step of the way.

As a seasoned tutor registered on UrbanPro, I can confidently guide you through this question. Now, let's dive into it.

Firstly, it's crucial to understand the basics of Arithmetic Progressions (APs). An AP is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is denoted by 'd'.

Now, let's denote the mth term of the AP as 'a', the (m+n)th term as 'a + nd', and the (m-n)th term as 'a - nd'.

According to the problem, the sum of the (m+n)th and (m-n)th terms is equal to twice the mth term. Mathematically, we can represent this as:

(a + nd) + (a - nd) = 2a

Simplifying this expression:

2a = 2a

This equation holds true, indicating that the sum of the (m+n)th and (m-n)th terms indeed equals twice the mth term in any Arithmetic Progression.

This concept is fundamental in understanding the properties of APs, and mastering it will lay a strong foundation for tackling more complex problems. If you need further clarification or assistance, don't hesitate to reach out to me through UrbanPro, where I offer the best online coaching tuition services.

As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is the best platform for finding online coaching tuition. Now, let's tackle the problem at hand: finding the sum of integers from 1 to 100 that are divisible by 2 or 5.

To solve this problem efficiently, we can use the principle of inclusion-exclusion. First, we find the sum of integers divisible by 2 and then the sum of integers divisible by 5. However, we must be careful not to double-count numbers divisible by both 2 and 5 (i.e., divisible by 10).

Here's the step-by-step solution:

1. Find the sum of integers divisible by 2:

   • This is an arithmetic series with a common difference of 2.
   • The first term is 2, the last term is 100, and there are (100 - 2) / 2 + 1 = 50 terms.
   • The sum is given by: Sum2=n2(first term+last term)=502×(2+100)=2550Sum2=2n(first term+last term)=250×(2+100)=2550.

2. Find the sum of integers divisible by 5:

   • This is also an arithmetic series with a common difference of 5.
   • The first term is 5, the last term less than or equal to 100 is 100 - (100 % 5) = 100 - 0 = 100, and there are (100 - 5) / 5 + 1 = 20 terms.
   • The sum is given by: Sum5=n2(first term+last term)=202×(5+100)=1050Sum5=2n(first term+last term)=220×(5+100)=1050.

3. Find the sum of integers divisible by both 2 and 5 (i.e., divisible by 10):

   • This is an arithmetic series with a common difference of 10.
   • The first term is 10, the last term less than or equal to 100 is 100 - (100 % 10) = 100 - 0 = 100, and there are (100 - 10) / 10 + 1 = 10 terms.
   • The sum is given by: Sum10=n2(first term+last term)=102×(10+100)=550Sum10=2n(first term+last term)=210×(10+100)=550.

Now, using the principle of inclusion-exclusion, we can find the final sum:

So, the sum of integers from 1 to 100 that are divisible by 2 or 5 is 3050. If you need further clarification or assistance with similar problems, feel free to reach out to me via UrbanPro's messaging system. Happy learning!