Learn Unit-II: Algebra with Free Lessons & Tips

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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.

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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

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!

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.

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:

• aa as the number of frames built on the first day (5 frames)
• dd as the common difference (the increase in frames built each day, which is 2 frames)
• nn as the number of days it takes to finish the job
• SS as the total number of frames built (192 frames)

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.

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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:

• The common difference of the AP as 'd'.
• The ath term as 'a'.
• The bth term as 'a + (b - 1)d' (since it's (b - 1) terms ahead of 'a').
• The cth term as 'a + (c - 1)d'.
• The dth term as 'a + (d - 1)d'.

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.