Question

Let $S_n$ and $s_n$ denotes the sum of first n terms of two different AP's whose $n^{th}$ term is $T_n$ and $t_n$ respectively. If $\frac{s_n}{S_n} = \frac{3n-13}{7n+13}$ then select the correct option(s)

• A. $\frac{t_{10}}{T_9} = \frac{3}{5}$
• B. $\frac{t_{10}}{T_9} = \frac{1}{3}$
• C. $\frac{s_n}{S_{2n}} = \frac{3n-13}{14n+13}$
• D. $\frac{s_n}{S_{2n}} = \frac{3n+13}{28n-26}$

Answer 1

Answer: (B)

$\frac{t_{10}}{T_9} = \frac{1}{3}$

Answer 2

The problem provides the ratio of the sum of the first n terms of two different arithmetic progressions (APs), $s_n$ and $S_n$. Let the first AP have first term $a_1$ and common difference $d_1$. Its $n^{th}$ term is $t_n$. Let the second AP have first term $a_2$ and common difference $d_2$. Its $n^{th}$ term is $T_n$.

The sum of the first n terms of an AP is given by $S_k = \frac{k}{2}[2a + (k-1)d]$. So, $s_n = \frac{n}{2}[2a_1 + (n-1)d_1]$ and $S_n = \frac{n}{2}[2a_2 + (n-1)d_2]$.

The given ratio is $\frac{s_n}{S_n} = \frac{3n-13}{7n+13}$. Substituting the sum formulas:

$$\frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{3n-13}{7n+13}$$ $$\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{3n-13}{7n+13}$$

To find the values of $a_1, d_1, a_2, d_2$, we can equate the expressions. Let's assume there exists a constant $C$ such that: $2a_1 + (n-1)d_1 = C(3n-13)$ $2a_2 + (n-1)d_2 = C(7n+13)$

Expanding the left side of the first equation: $2a_1 - d_1 + nd_1$. Comparing coefficients of $n$ and constant terms with $3Cn - 13C$: For coefficient of $n$: $d_1 = 3C$ For constant term: $2a_1 - d_1 = -13C$ Substitute $d_1 = 3C$: $2a_1 - 3C = -13C \implies 2a_1 = -10C \implies a_1 = -5C$.

Expanding the left side of the second equation: $2a_2 - d_2 + nd_2$. Comparing coefficients of $n$ and constant terms with $7Cn + 13C$: For coefficient of $n$: $d_2 = 7C$ For constant term: $2a_2 - d_2 = 13C$ Substitute $d_2 = 7C$: $2a_2 - 7C = 13C \implies 2a_2 = 20C \implies a_2 = 10C$.

Now we have the first term and common difference for both APs in terms of $C$: First AP: $a_1 = -5C$, $d_1 = 3C$. Second AP: $a_2 = 10C$, $d_2 = 7C$.

We need to evaluate the options. Let's start with the terms $t_{10}$ and $T_9$. The $k^{th}$ term of an AP is given by $a_k = a + (k-1)d$. For the first AP, $t_{10} = a_1 + (10-1)d_1 = a_1 + 9d_1$. Substitute $a_1 = -5C$ and $d_1 = 3C$: $t_{10} = -5C + 9(3C) = -5C + 27C = 22C$.

For the second AP, $T_9 = a_2 + (9-1)d_2 = a_2 + 8d_2$. Substitute $a_2 = 10C$ and $d_2 = 7C$: $T_9 = 10C + 8(7C) = 10C + 56C = 66C$.

Now, calculate the ratio $\frac{t_{10}}{T_9}$:

$$\frac{t_{10}}{T_9} = \frac{22C}{66C} = \frac{1}{3}$$

This matches the second option.

Calculate $\frac{s_n}{S_{2n}}$: $s_n = \frac{n}{2}[2a_1 + (n-1)d_1] = \frac{n}{2}[2(-5C) + (n-1)(3C)] = \frac{n}{2}[-10C + 3nC - 3C] = \frac{nC}{2}(3n-13)$.

$S_{2n} = \frac{2n}{2}[2a_2 + (2n-1)d_2] = n[2(10C) + (2n-1)(7C)] = n[20C + 14nC - 7C] = nC(13+14n)$.

Now, find the ratio $\frac{s_n}{S_{2n}}$:

$$\frac{s_n}{S_{2n}} = \frac{\frac{nC}{2}(3n-13)}{nC(14n+13)} = \frac{3n-13}{2(14n+13)} = \frac{3n-13}{28n+26}$$