Question

Let $a_{1,}{a}_2,a_3,..$ be terms of an A.P. whose common difference is an integer and $S_n=a_1+a_2+...+a_n$. If $a_1=1,a_n=300$ and $15\le n\le 50,$then the ordered pair $\left(S_{n-4},a_{n-4}\right)$ is equal to:

• A. $\left(2480,248\right)$
• B. $\left(2480,249\right)$
• C. $\left(2490,249\right)$
• D. $\left(2490,248\right)$

Answer 1

The correct option is D

$\left(2490,248\right)$

Explanation for correct option

Step 1: Determine the value of $d$

An arithmetic progression (AP) is a progression in which the difference between two consecutive terms is constant.

$n^{th}$ term of an A.P. is $a_n=a_1+\left(n-1\right)d$ where $a_1=\text{first term},d=\text{common difference}$

Given,$a_1=1,a_n=300$

$$\begin{gathered} a_n=a_1+\left(n-1\right)d \\ 300=1+\left(n-1\right)d \\ 299=\left(n-1\right)d \\ n-1=\frac{299}{d} \end{gathered}$$

Now, given that common difference is integer and also $n-1$ will be an integer.

So, $d=23\text{or}13$

If $d=23$,

$$\begin{gathered} \Rightarrow n-1=13 \\ \Rightarrow n=14 \end{gathered}$$

It is impossible as, $15\le n\le 50$.

Again If $d=13$

$$\begin{gathered} \Rightarrow n-1=23 \\ \Rightarrow n=24 \end{gathered}$$

It is possible. since $15\le n\le 50$.

Step 2: To find $\left(S_{n-4},a_{n-4}\right)$

Sum of $n$ terms of AP is given by: $S_n=\frac{n}{2}\left[2a_1+\left(n-1\right)d\right]$ where $a_1=\text{first term},d=\text{common difference}$.

$$\begin{gathered} S_{24-4}=\frac{24-4}{2}\left[2\times 1+\left(24-4-1\right)13\right]\left[\because d=13,n=24,a_1=1\right] \\ {S}_{20}=\frac{20}{2}\left[2+19\times 13\right] \\ =2490 \end{gathered}$$

$$\begin{gathered} a_{n-4}=a_1+\left(n-4-1\right)d \\ {a}_{24-4}=1+\left(24-4-1\right)13\left[\because d=13,n=24,a_1=1\right] \\ {a}_{20}=248 \end{gathered}$$

Therefore, the ordered pair $\left(S_{n-4},a_{n-4}\right)$ is equal to $(2490,248)$.

Hence, option (D) is the correct answer.