Question

(i) The sum of n terms of two arithmetic series are in the ratio of (7n + 1) : (4n + 27). Find the ratio of their nth term.

(ii) In an AP of which 'a' is the Ist term, if the sum of the Ist p terms is equal to zero, show that the sum of the next q terms is $-\left(\frac{aq(p+q)}{p-1}\right)$

Answer 1

(i) (14n - 6) : (8n + 23) (ii) Proof shown above

Answer 2

The problem consists of two independent parts related to Arithmetic Progressions (AP).

Part (i): Ratio of nth term of two APs

Let the first arithmetic series have first term $a_1$ and common difference $d_1$.
Let the second arithmetic series have first term $a_2$ and common difference $d_2$.

The sum of $N$ terms of an AP is given by the formula $S_N = \frac{N}{2}[2a + (N-1)d]$.
The ratio of the sum of $N$ terms of the two APs is given as: $\frac{S_{N,1}}{S_{N,2}} = \frac{\frac{N}{2}[2a_1 + (N-1)d_1]}{\frac{N}{2}[2a_2 + (N-1)d_2]} = \frac{2a_1 + (N-1)d_1}{2a_2 + (N-1)d_2}$ We are given that this ratio is $(7N + 1) : (4N + 27)$.
So, $\frac{2a_1 + (N-1)d_1}{2a_2 + (N-1)d_2} = \frac{7N + 1}{4N + 27} \quad (*)$

We need to find the ratio of their $n^{th}$ terms. The $n^{th}$ term of an AP is given by $T_n = a + (n-1)d$.
So, we need to find $\frac{T_{n,1}}{T_{n,2}} = \frac{a_1 + (n-1)d_1}{a_2 + (n-1)d_2}$.

To transform the expression in $(*)$ into the desired form, we observe that:
$2a + (N-1)d = 2[a + \frac{N-1}{2}d]$.
We want the term $a_1 + (n-1)d_1$.
So, we need to choose $N$ such that $\frac{N-1}{2} = n-1$.
This implies $N-1 = 2(n-1)$, which gives $N = 2(n-1) + 1 = 2n - 2 + 1 = 2n - 1$.

Substitute $N = 2n-1$ into equation $(*)$: $\frac{2a_1 + ((2n-1)-1)d_1}{2a_2 + ((2n-1)-1)d_2} = \frac{7(2n-1) + 1}{4(2n-1) + 27}$ $\frac{2a_1 + (2n-2)d_1}{2a_2 + (2n-2)d_2} = \frac{14n - 7 + 1}{8n - 4 + 27}$ $\frac{2[a_1 + (n-1)d_1]}{2[a_2 + (n-1)d_2]} = \frac{14n - 6}{8n + 23}$ $\frac{a_1 + (n-1)d_1}{a_2 + (n-1)d_2} = \frac{14n - 6}{8n + 23}$ Thus, the ratio of their $n^{th}$ term is $(14n - 6) : (8n + 23)$.

Part (ii): Sum of next q terms

Let the AP have first term 'a' and common difference 'd'.

Given that the sum of the first $p$ terms is zero:
$S_p = 0$
Using the formula $S_p = \frac{p}{2}[2a + (p-1)d]$, we have: $\frac{p}{2}[2a + (p-1)d] = 0$ Since $p$ is the number of terms, $p \ne 0$. Therefore, $2a + (p-1)d = 0$ From this, we can express $d$ in terms of $a$ and $p$. Assuming $p \ne 1$ (as division by $p-1$ occurs): $(p-1)d = -2a$ $d = \frac{-2a}{p-1}$

The sum of the next $q$ terms refers to the terms from $(p+1)^{th}$ to $(p+q)^{th}$.
This sum can be calculated as the sum of the first $(p+q)$ terms minus the sum of the first $p$ terms:
Sum of next $q$ terms $= S_{p+q} - S_p$.
Since $S_p = 0$, the sum of the next $q$ terms is $S_{p+q}$.

Now, let's calculate $S_{p+q}$: $S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d]$ Substitute the expression for $d = \frac{-2a}{p-1}$: $S_{p+q} = \frac{p+q}{2}\left[2a + (p+q-1)\left(\frac{-2a}{p-1}\right)\right]$ Factor out $2a$: $S_{p+q} = \frac{p+q}{2} \cdot 2a \left[1 - \frac{p+q-1}{p-1}\right]$ $S_{p+q} = a(p+q) \left[\frac{(p-1) - (p+q-1)}{p-1}\right]$ $S_{p+q} = a(p+q) \left[\frac{p-1 - p - q + 1}{p-1}\right]$ $S_{p+q} = a(p+q) \left[\frac{-q}{p-1}\right]$ $S_{p+q} = -\frac{aq(p+q)}{p-1}$ This shows that the sum of the next $q$ terms is $-\left(\frac{aq(p+q)}{p-1}\right)$.

The final answer is $\boxed{\text{(i) (14n - 6) : (8n + 23) (ii) Proof shown above}}$.

Explanation of the solution:

(i) To find the ratio of the $n^{th}$ terms from the ratio of sums of $N$ terms, we replace $N$ with $2n-1$ in the given ratio expression. This transformation converts the sum formula $2a+(N-1)d$ into $2[a+(n-1)d]$, which simplifies to the ratio of $n^{th}$ terms.
(ii) Given $S_p=0$, we derived $d = \frac{-2a}{p-1}$. The sum of the next $q$ terms is $S_{p+q}-S_p = S_{p+q}$. We substituted the expression for $d$ into the formula for $S_{p+q}$ and simplified to obtain the required result.

Answer:

(i) The ratio of their nth term is $\frac{14n - 6}{8n + 23}$.
(ii) Proof shown above.