Question

If the sum of the first $2n$terms of the A.P. $2,5,8,...$is equal to the sum of the first $n$terms of the A.P. $57,59,61,...$, then $n$equals to

• A. $10$
• B. $12$
• C. $11$
• D. $13$

Answer 1

The correct option is C

$11$

Arithmetic progression (A.P):

Arithmetic progression is defined as a series in which the difference between any two consecutive terms is constant throughout the series. The constant difference is called the common difference.

Sum of $n$terms of an A.P.:

The sum of the first $n$terms of an A.P with the first term $a$ and the common difference $d$is given by

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

The explanation for the correct option: (Option C)

Step 1: Finding $S_{2n}$

The sum of the first $2n$terms of the A.P. $2,5,8,...$

Here the first term is $2$ and common difference $3.$

[As, $5-2=3,8-5=3$]

$$\begin{gathered} S_{2n}=\frac{2n}{2}[2\times 2+(2n-1)\times 3] \\ =n[4+6n-3] \\ =n[1+6n] \end{gathered}$$

Step 1: Finding $S_n$

The sum of the first $n$terms of the A.P. $57,59,61,...$

Here the first term is $57$and the common difference is $2.$

[As, $59-57=2,61-59=2$]

Then,

$$\begin{gathered} S_n=\frac{n}{2}[2\times 57+(n-1)\times 2] \\ =\frac{n}{2}[114+2n-2] \\ =\frac{n}{2}[112+2n] \\ =n[56+n] \end{gathered}$$

Step 3: Finding n

According to the question,

$$\begin{gathered} S_{2n}=S_n \\ \Rightarrow n[1+6n]=n[56+n] \\ \Rightarrow 1+6n=56+n \\ \Rightarrow 6n-n=56-1 \\ \Rightarrow 5n=55 \\ \Rightarrow n=11 \end{gathered}$$

Conclusion:

Hence the value of $nis11.$(Option C)