The sum of $n$ terms of two arithmetic progressions are in the ratio $(9n + 5) : (5n + 21)$. Find the ratio of their $15^{th}$ terms.
133 : 83
Arithmetic Progression: Ratio of Terms from Sum Ratio
This problem asks us to find the ratio of the $15^{th}$ terms of two arithmetic progressions (APs). We are given information about the ratio of the sums of their first n terms. To solve this, we will use the standard formulas for arithmetic progressions.
Understanding Arithmetic Progression Formulas
Let's denote the first arithmetic progression as AP1 and the second as AP2.
- For AP1, let the first term be $a$ and the common difference be $d$.
- For AP2, let the first term be $a'$ and the common difference be $d'$.
The formula for the sum of the first n terms of an AP is:
$$ S_n = \frac{n}{2}[2a + (n-1)d] $$And the formula for the kth term of an AP is:
$$ a_k = a + (k-1)d $$Relating the Ratio of Sums to the Ratio of Terms
The problem states that the ratio of the sums of the first n terms of the two APs is:
$$ \frac{S_n}{S'_n} = \frac{9n + 5}{5n + 21} $$Using the sum formula, we can write the ratio of sums as:
$$ \frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a + (n-1)d]}{\frac{n}{2}[2a' + (n-1)d']} = \frac{2a + (n-1)d}{2a' + (n-1)d'} $$Equating the two expressions for the ratio of sums:
$$ \frac{2a + (n-1)d}{2a' + (n-1)d'} = \frac{9n + 5}{5n + 21} $$We need to find the ratio of the $15^{th}$ terms:
$$ \frac{a_{15}}{a'_{15}} = \frac{a + (15-1)d}{a' + (15-1)d'} = \frac{a + 14d}{a' + 14d'} $$Notice that the expression for the $k^{th}$ term ratio, $\frac{a + (k-1)d}{a' + (k-1)d'}$, can be obtained from the sum ratio expression $\frac{2a + (n-1)d}{2a' + (n-1)d'}$ if we relate $n-1$ to $k-1$. Specifically, we need $n-1 = 2(k-1)$.
For the $15^{th}$ term, we have $k=15$, so $k-1 = 14$. We set:
$$ n-1 = 2(15-1) = 2 \times 14 = 28 $$This implies $n = 28 + 1 = 29$.
Substituting $n=29$ into the ratio of sums formula effectively gives us the ratio of the $15^{th}$ terms:
$$ \frac{2a + (29-1)d}{2a' + (29-1)d'} = \frac{2a + 28d}{2a' + 28d'} = \frac{2(a + 14d)}{2(a' + 14d')} = \frac{a + 14d}{a' + 14d'} = \frac{a_{15}}{a'_{15}} $$Calculation of the 15th Terms Ratio
Now we substitute $n=29$ into the given expression for the ratio of sums:
$$ \frac{a_{15}}{a'_{15}} = \frac{9(29) + 5}{5(29) + 21} $$Let's calculate the values:
- Numerator: $9 \times 29 + 5 = 261 + 5 = 266$
- Denominator: $5 \times 29 + 21 = 145 + 21 = 166$
So, the ratio of the $15^{th}$ terms is:
$$ \frac{a_{15}}{a'_{15}} = \frac{266}{166} $$To simplify the ratio, we find the greatest common divisor of 266 and 166, which is 2. Divide both numbers by 2:
$$ \frac{266 \div 2}{166 \div 2} = \frac{133}{83} $$Thus, the ratio of the $15^{th}$ terms is 133 : 83.
Final Answer
The ratio of the $15^{th}$ terms of the two arithmetic progressions is 133:83.
Important Questions from Arithmetic Progressions
The nth term of an A.P is \(\frac{{3 + {\rm{n}}}}{4}\) , then the sum of first 105 terms is
What is the sum of n terms of the series \(\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} + \ldots ?\)
Find the sum of all even numbers between 1 to 100.
Which of the following disciplines studies human populations mostly with respect to their size, their structure and their development?
What is the sum of all two digit odd numbers?
