Question

For an AP of 14 terms, find the option which gives the value equal to $a_4+a_{11}$.

• A. $a_3+a_8$
• B. $a_6+a_9$
• C. $a_9+a_7$
• D. $a_5+a_7$

Answer 1

The correct option is B $a_6+a_9$

For an AP, $a_1,a_2,a_3,.......,a_n$

$a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}$ and so on.

By observing we get to know that sum of terms are equal , if its sum of position value are equal.

Here in this AP , $a_1+a_{14}=a_2+a_{13}=a_3+a_{12}$ and so on.

For, $a_4+a_{11}$ , sum of position values $=4+11=15$

From the options:
$\left(i\right)$ $a_3+a_8$, sum of position values $=3+8=11$

$\left(ii\right)$ $a_6+a_9$, sum of position values $=6+9=15$

$\left(iii\right)$ $a_9+a_7$, sum of position values $=7+9=16$

$\left(iv\right)$ $a_5+a_7$, sum of position values $=5+7=12$

The sum of position values of $a_4+a_{11}$ is equal to sum of position values of $a_6+a_9$
$\therefore a_4+a_{11}=a_6+a_9$