Problem 66
Question
The ratio between the sum of \(n\) term of two A.P.'s is \(3 n+8: 7 n+15\). Find the ratio between their 12 th terms. Also find the ratio of their common difference.
Step-by-Step Solution
Verified Answer
The ratio of their 12th terms is \( \frac{32}{99} \). The ratio of their common differences is \( \frac{14}{29} \).
1Step 1: Understand the given
We know that the sum of \(n\) terms of an AP is given by \(\frac{n}{2}[2a + (n-1)d]\) where \(a\) is the first term and \(d\) is the common difference. Here, two different APs are given and for each, we can say that the sum \(S\) of their \(n\) terms can be referred to as \(S_1\) and \(S_2\) respectively. Therefore, the given ratio is \(\frac{S_1}{S_2} = \frac{3n+8}{7n+15}\).
2Step 2: Obtain ratio of terms
Substitute the sum \(S\) of an AP in the given ratio. We have, \(\frac{\frac{n}{2}[2a_1 + (n - 1)d_1]}{\frac{n}{2}[2a_2 + (n - 1)d_2]} = \frac{3n + 8}{7n + 15}\). On simplification, we get \(\frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} = \frac{3n + 8}{7n + 15}\). Now, replace \(n\) with \(12\) to get the ratio of their 12th terms, that is \(T_1/T_2\). Thus we now have, \(T_1/T_2 = \frac{2a_1 + 11d_1}{2a_2 + 11d_2} = \frac{3*12 + 8}{7*12 + 15}\).
3Step 3: Obtain ratio of common differences
For the ratio of the common difference, put \(n = 2\). Putting \(n = 2\) in the given ratio, we can get the ratio of the common difference as \(d_1/d_2 = \frac{3*2 + 8}{7*2 + 15}\).
Key Concepts
Arithmetic ProgressionSum of n terms of APCommon difference of AP
Arithmetic Progression
Arithmetic Progression (AP) is a sequence of numbers in which the difference of any two successive members is a constant, known as the common difference. An AP series resembles the formula:
\(a, a+d, a+2d, a+3d, ...\), where \(a\) represents the first term and \(d\) is the common difference. The AP is fundamental in understanding patterns within sequences and forms the basis for various calculations in algebra.
For instance, the sequence 4, 7, 10, 13 is an AP because the difference between consecutive terms is always 3. The constant nature of this difference is what defines AP. Understanding this is crucial as it allows us to predict later terms in a series and calculate the sum of the series, which can be critical in solving many mathematical and real-world problems.
\(a, a+d, a+2d, a+3d, ...\), where \(a\) represents the first term and \(d\) is the common difference. The AP is fundamental in understanding patterns within sequences and forms the basis for various calculations in algebra.
For instance, the sequence 4, 7, 10, 13 is an AP because the difference between consecutive terms is always 3. The constant nature of this difference is what defines AP. Understanding this is crucial as it allows us to predict later terms in a series and calculate the sum of the series, which can be critical in solving many mathematical and real-world problems.
Sum of n terms of AP
The sum of the first \(n\) terms of an Arithmetic Progression can be calculated using the formula: \[S_n = \frac{n}{2}(2a + (n-1)d)\]
Here, \(S_n\) represents the sum of the first \(n\) terms, while \(a\) stands for the first term and \(d\) for the common difference. This formula is derived by multiplying the average of the first and last term by the number of terms.
For example, to calculate the sum of the first 5 terms of an AP that starts with 2 and has a common difference of 3, we use \(a = 2\), \(d = 3\), and \(n = 5\) to get: \[S_5 = \frac{5}{2}(2\times2 + (5-1)\times3) = \frac{5}{2}(4 + 12) = \frac{5}{2}\times16 = 40\].
It's a versatile formula that can be applied in various scenarios involving AP, such as finding the total distance traveled in equal increments of speed, determining savings over time with constant contributions, and many more practical situations.
Here, \(S_n\) represents the sum of the first \(n\) terms, while \(a\) stands for the first term and \(d\) for the common difference. This formula is derived by multiplying the average of the first and last term by the number of terms.
For example, to calculate the sum of the first 5 terms of an AP that starts with 2 and has a common difference of 3, we use \(a = 2\), \(d = 3\), and \(n = 5\) to get: \[S_5 = \frac{5}{2}(2\times2 + (5-1)\times3) = \frac{5}{2}(4 + 12) = \frac{5}{2}\times16 = 40\].
It's a versatile formula that can be applied in various scenarios involving AP, such as finding the total distance traveled in equal increments of speed, determining savings over time with constant contributions, and many more practical situations.
Common difference of AP
The common difference in an Arithmetic Progression is the uniform amount by which consecutive terms increase or decrease. It is represented by \(d\) and found by subtracting a term from the one following it: \[d = a_{n} - a_{n-1}\]
The common difference is a critical element because it defines the particular nature of the progression. Whether the common difference is positive, negative, or zero, it shapes the sequence into ascending, descending, or constant, respectively.
For instance, if we consider a sequence such as 5, 9, 13, 17, ... the common difference is \(9 - 5 = 4\).
The common difference is a critical element because it defines the particular nature of the progression. Whether the common difference is positive, negative, or zero, it shapes the sequence into ascending, descending, or constant, respectively.
For instance, if we consider a sequence such as 5, 9, 13, 17, ... the common difference is \(9 - 5 = 4\).
Calculating Common Difference from the Ratio of Sums
In the context of the exercise, to find the ratio of their common differences, we can rearrange the given sum of term ratios. It is useful in comparing two sequences and understanding how their differences relate. This calculation is particularly important for problems where we have to find out how two different APs compare to each other in terms of growth or decay rate.Other exercises in this chapter
Problem 64
Show that the sum of an A.P. whose first term is \(a\), second term is \(b\) and the last term is \(c\) is equal to \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\)
View solution Problem 65
If the ratio of the sum of \(m\) term and \(n\) terms of an A.P. be \(m^{2}: n^{2}\), prove that the ratio of its \(m\) th and \(n\) th terms will be \(2 m-1: 2
View solution Problem 67
The ratio between the sum of \(n\) term of two A.P.'s is \(7 n+1: 4 n+27\). Find the ratio between their \(n\) th terms.
View solution Problem 68
The sum of the first \(n\) term of two A.P.'s are as \(3 n+5: 5 n-9\). Prove that their 4 th terms are equal.
View solution