Question

Let $x_1,x_2,\dots x_n$ be in an AP. $x_1+x_4+x_9+x_{11}+x_{20}+x_{22}+x_{27}+x_{30}=272$ then $x_1+x_3+\dots +x_{30}$ is equal to

• A. $1020$
• B. $1200$
• C. $716$
• D. $2720$

Answer 1

The correct option is A

$1020$

Explanation for the correct option:

Finding the sum of all terms:

Let $a$ be the first term and $d$ be the common difference of AP.

We have given that, $x_1+x_4+x_9+x_{11}+x_{20}+x_{22}+x_{27}+x_{30}=272$

since, $n^{th}$ term of an AP is calculated as $a_n=a+\left(n-1\right)d$ so, above equation becomes

$a+a+3d+a+8d+a+10d+a+19d+a+21d+a+26d+a+29d=272$

On simplifying above equation, we get

$$\begin{gathered} 8a+116d=272 \\ 2a+29d=68...\left(i\right) \end{gathered}$$

Since, there are total $n=30$ terms in the series so, the sum of and AP of $n=30$ is given by

$S_n=\left(\frac{n}{2}\right)\left(a_1+a_n\right)$

On substituting the values we get,

$$\begin{gathered} \Rightarrow x_1+x_3+\dots +x_{30}=\left(\frac{30}{2}\right)\left(a+\left(a+29d\right)\right) \\ \Rightarrow x_1+x_3+\dots +x_{30}=15\left(2a+29d\right) \end{gathered}$$

Using the value from equation $\left(i\right)$ we get,

$$\begin{gathered} \Rightarrow x_1+x_3+\dots +x_{30}=15\left(2a+29d\right) \\ \Rightarrow x_1+x_3+\dots +x_{30}=15\times 68 \\ \Rightarrow x_1+x_3+\dots +x_{30}=1020 \end{gathered}$$

So, the sum of the gievn AP $x_1+x_3+\dots +x_{30}$ is $1020$

Alternative solution:

Artithmetic mean of symmetrically placed A.P terms are always equal

$\begin{array}{rcl}\frac{x_1+x_4+x_9+x_{11}+x_{20}+x_{22}+x_{27}+x_{30}}{8}& =& \frac{sumofallterms}{30}\\ \Rightarrow x_1+x_2+x_3+......+x_{30}& =& \frac{30\times 272}{8}\\ & =& 1020\end{array}$

Hence, the correct answer is option (A).