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Answered on 14 Apr Learn Sequence and Series
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into solving the problem at hand.
Given that the sums of nn terms of two arithmetic progressions are in the ratio 5n+4:9n+65n+4:9n+6, we need to find the ratio of their 18th terms.
For an arithmetic progression, the sum of the first nn terms is given by Sn=n2[2a+(n−1)d]Sn=2n[2a+(n−1)d], where aa is the first term and dd is the common difference.
So, for the first arithmetic progression, let's denote its first term as a1a1 and common difference as d1d1, and for the second arithmetic progression, let's denote its first term as a2a2 and common difference as d2d2.
Given that the sum of nn terms of the first arithmetic progression is 5n+45n+4, we have: n2[2a1+(n−1)d1]=5n+42n[2a1+(n−1)d1]=5n+4 Similarly, for the second arithmetic progression with sum 9n+69n+6, we have: n2[2a2+(n−1)d2]=9n+62n[2a2+(n−1)d2]=9n+6
We are given that these sums are in the ratio 5n+4:9n+65n+4:9n+6. Therefore, we can write: 5n+49n+6=n2[2a1+(n−1)d1]n2[2a2+(n−1)d2]9n+65n+4=2n[2a2+(n−1)d2]2n[2a1+(n−1)d1]
Now, simplifying this equation and solving for a2a1a1a2 will give us the ratio of their 18th terms. But first, let's solve for a1a1 and a2a2 by eliminating nn from the given ratios.
5n+49n+6=5n+49n+69n+65n+4=9n+65n+4 n2[2a1+(n−1)d1]n2[2a2+(n−1)d2]=5n+49n+62n[2a2+(n−1)d2]2n[2a1+(n−1)d1]=9n+65n+4 2a1+(n−1)d12a2+(n−1)d2=5n+49n+62a2+(n−1)d22a1+(n−1)d1=9n+65n+4
Now, let's simplify this expression. We'll replace nn with 18, as we are interested in the ratio of their 18th terms. Then, we'll solve for a2a1a1a2 to find the ratio.
Answered on 14 Apr Learn Sequence and Series
Nazia Khanum
Sure, as a experienced tutor registered on UrbanPro, I can help you with that. UrbanPro is indeed one of the best platforms for online coaching and tuition.
To insert five numbers between 8 and 26 such that the resulting sequence forms an Arithmetic Progression (A.P.), we need to find the common difference first. The common difference (d) of an A.P. is calculated by subtracting the first term from the second term or vice versa.
Let's denote the first term of our A.P. as a1=8a1=8 and the last term as an=26an=26.
The formula to find the nth term of an A.P. is an=a1+(n−1)⋅dan=a1+(n−1)⋅d.
Given an=26an=26 and a1=8a1=8, we can find the common difference (d).
26=8+(n−1)⋅d26=8+(n−1)⋅d
18=(n−1)⋅d18=(n−1)⋅d
Now, let's choose a value for nn, say n=6n=6 (because we need to insert 5 numbers between 8 and 26).
18=(6−1)⋅d18=(6−1)⋅d 18=5d18=5d d=185=3.6d=518=3.6
Now, we can find the numbers by adding the common difference to each preceding term.
a2=a1+d=8+3.6=11.6a2=a1+d=8+3.6=11.6 a3=a2+d=11.6+3.6=15.2a3=a2+d=11.6+3.6=15.2 a4=a3+d=15.2+3.6=18.8a4=a3+d=15.2+3.6=18.8 a5=a4+d=18.8+3.6=22.4a5=a4+d=18.8+3.6=22.4 a6=a5+d=22.4+3.6=26a6=a5+d=22.4+3.6=26
So, the resulting sequence forming an A.P. with 5 numbers inserted between 8 and 26 is:
8,11.6,15.2,18.8,22.4,268,11.6,15.2,18.8,22.4,26
Answered on 14 Apr Learn Sequence and Series
Nazia Khanum
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:
From the 5th and 8th terms: q=ar7q=ar7
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.
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Answered on 14 Apr Learn Sequence and Series
Nazia Khanum
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.
Answered on 14 Apr Learn Sequence and Series
Nazia Khanum
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:
Find the sum of integers divisible by 2:
Find the sum of integers divisible by 5:
Find the sum of integers divisible by both 2 and 5 (i.e., divisible by 10):
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!
Answered on 14 Apr Learn Sequence and Series
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's tackle the math problem at hand.
We're given that the sum of three numbers in an arithmetic progression (AP) is 24, and their product is 44. To find the three numbers, let's denote the common difference of the AP as 'd' and the middle number as 'b'.
According to the formula for the sum of an arithmetic progression, the sum of three numbers in an AP is given by: Sum=n2(2a+(n−1)d)Sum=2n(2a+(n−1)d) where 'n' is the number of terms, 'a' is the first term, and 'd' is the common difference.
Given that the sum of the three numbers is 24, and they form an arithmetic progression, we can set up the equation: 24=32(2a+2d)24=23(2a+2d) 24=3(a+d)24=3(a+d) 8=a+d8=a+d
Now, according to the formula for the product of three numbers in an arithmetic progression, it is given by: Product=abcProduct=abc Given that the product is 44, we have: 44=a(b)(c)44=a(b)(c) 44=a(b)(a+2d)44=a(b)(a+2d) 44=ab2+2ad44=ab2+2ad
We already know that a+d=8a+d=8, so we can substitute a=8−da=8−d into the equation for the product: 44=(8−d)b2+2d(8−d)44=(8−d)b2+2d(8−d) 44=8b2−db2+16d−2d244=8b2−db2+16d−2d2 0=8b2−db2+16d−2d2−440=8b2−db2+16d−2d2−44 0=(8−d)b2+2(8−d)d−440=(8−d)b2+2(8−d)d−44
Now, we need to find the values of 'b' and 'd' that satisfy this equation. Let's try different values of 'b' and 'd' that make sense within the given constraints.
Once we find the values of 'b' and 'd', we can easily find the first term 'a' and then the other two numbers in the arithmetic progression. This problem involves some algebraic manipulation and solving quadratic equations. If you need further assistance, feel free to ask!
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Answered on 14 Apr Learn Sequence and Series
Nazia Khanum
Certainly! Let's approach this problem systematically. Since we are dealing with a geometric progression (GP), let's represent the three numbers as arar, ar2ar2, and ar3ar3, where aa is the first term and rr is the common ratio.
We're given that the product of these three numbers is 1728, so we can form the equation:
(ar)(ar2)(ar3)=1728(ar)(ar2)(ar3)=1728
a3r6=1728a3r6=1728
Similarly, we're also given that the sum of these three numbers is 38, so:
ar+ar2+ar3=38ar+ar2+ar3=38
Now, let's solve these equations step by step. Firstly, let's rewrite the product equation:
a3r6=1728a3r6=1728
a3r6=26×33a3r6=26×33
(ar)3×(ar2)3=26×33(ar)3×(ar2)3=26×33
Now, we can see that ar=2ar=2 and ar2=3ar2=3, because 2 and 3 are the prime factors of 1728 and they are consecutive powers. So, let's solve for aa and rr:
From ar=2ar=2, we get a=2ra=r2
From ar2=3ar2=3, we get a=3r2a=r23
Equating these expressions for aa:
2r=3r2r2=r23
2r=32r=3
r=32r=23
Now that we have found rr, we can find aa:
ar=2ar=2
a×32=2a×23=2
a=43a=34
So, the three numbers are 4334, 2, and 3.
Therefore, as an experienced tutor registered on UrbanPro, I have demonstrated how to find three numbers in GP whose product is 1728 and sum is 38. UrbanPro is indeed a great platform for seeking online coaching and tuition services, where experienced tutors can provide comprehensive assistance in various subjects.
Answered on 14 Apr Learn Sequence and Series
Nazia Khanum
As an experienced tutor registered on UrbanPro, I often encounter intriguing mathematical problems like this one. Let's tackle it together!
In this scenario, we're dealing with a situation where the number of frames built each day forms an arithmetic progression. To find out how many days it took for the carpenter to complete the job, we need to sum the progression until we reach 192 frames.
Let's denote:
We know that the sum of an arithmetic progression is given by the formula: S=n2×(2a+(n−1)d)S=2n×(2a+(n−1)d)
Substituting the given values: 192=n2×(2×5+(n−1)×2)192=2n×(2×5+(n−1)×2)
Now, let's solve for nn: 192=n2×(10+2n−2)192=2n×(10+2n−2) 192=n2×(2n+8)192=2n×(2n+8) 192=n2×2×(n+4)192=2n×2×(n+4) 192=n×(n+4)192=n×(n+4)
Now, we need to find two consecutive integers whose product is 192. Upon calculation, we find that 12 and 16 are those integers.
So, n=12n=12. Therefore, it took the carpenter 12 days to finish the job.
UrbanPro is indeed a fantastic platform for honing mathematical skills like these, providing a wealth of resources and knowledgeable tutors to guide students through such challenges.
Answered on 14 Apr Learn Sequence and Series
Nazia Khanum
Certainly! As an experienced tutor registered on UrbanPro, I'd like to guide you through this problem using the properties of an arithmetic progression (AP).
Firstly, it's important to understand that in an arithmetic progression, each term is obtained by adding a fixed number (called the common difference) to the previous term.
Given: The arithmetic mean (AM) between the ath and bth terms of an AP is equal to the AM between the cth and dth terms.
Let's denote:
Now, we know that the arithmetic mean (AM) between two terms is the average of those terms.
So, the AM between the ath and bth terms is: a+(a+(b−1)d)2=2a+bd−d22a+(a+(b−1)d)=22a+bd−d
Similarly, the AM between the cth and dth terms is: a+(a+(c−1)d)2=2a+cd−d22a+(a+(c−1)d)=22a+cd−d
Given that these two AMs are equal, we have: 2a+bd−d2=2a+cd−d222a+bd−d=22a+cd−d
Now, let's simplify this equation: 2a+bd−d=2a+cd−d2a+bd−d=2a+cd−d bd=cdbd=cd
Now, subtracting 'a' from both sides: b−a=c−ab−a=c−a
Adding 'a' to both sides: a+b=a+ca+b=a+c
Finally, subtracting 'a' from both sides again: a+b=c+da+b=c+d
Thus, we have proven that if the arithmetic mean (AM) between the ath and bth terms of an arithmetic progression (AP) is equal to the AM between the cth and dth term, then a+b=c+da+b=c+d.
This demonstrates the fundamental property of arithmetic progressions, and understanding it can aid in solving various problems in mathematics. If you need further clarification or assistance with similar problems, UrbanPro is an excellent platform to find skilled tutors who can provide personalized guidance.
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Answered on 14 Apr Learn Permutations and Combinations
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I'd be delighted to assist you with your query. When a coin is tossed six times, we're essentially dealing with a sequence of events, each event having two possible outcomes: either heads or tails. To calculate the total number of possible outcomes, we can use the fundamental principle of counting, also known as the multiplication principle.
For each toss, there are 2 possible outcomes (heads or tails). Since there are 6 tosses in total, we multiply the number of outcomes for each toss together:
2 * 2 * 2 * 2 * 2 * 2 = 64
So, there are 64 possible outcomes when a coin is tossed 6 times. This understanding is fundamental for probability calculations and can be applied to various scenarios in probability and statistics. If you have any further questions or need clarification, feel free to ask. And remember, UrbanPro is a fantastic resource for finding experienced tutors like myself for all your academic needs.
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