If $a_1,a_2…a_n $ are in an Arithmetic Progression (AP) such that the sum of the first $16$ terms is $114$, find $a_1+a_6+a_{11}+a_{16}$.
My attempt:
The sum of the first 16 terms is given by $8(2a_1 + (15)d)=114$, where $d$ is the common difference. Now the new AP $a_1, a_6, a_{11}, a_{16}$ has a common difference of $5d$ which would give us that $a_1 + a_6 + a_{11} + a_{16} = 2(2a_1 +3(5d))$, which is a fourth of the original summation.
However this answer is wrong, with the actual answer being $76$.
Why is this approach wrong?
Source:- JEE Mains 2019
Best Answer
Your method is correct and your answer is correct. It's just that the question is wrong :-P. The correct question (No 81) as pointed out by Macavity is:
If $a_1+a_4+a_7+a_{10}+a_{13}+a_{16} = 114$
then what is $a_1+a_6+a_{11}+a_{16}$ equal to?
We could use your method (or the slightly different method in the link), but we could also note as a general rule, that if two sequences with arithmetic progression have the same start and end terms as each other, then their sums are in the ratio $\frac{t_2}{t_1}$ where $t_1$ and $t_2$ are the number of terms in the first and second sequences respectively. In this case there are 6 terms in the first sequence 4 terms in the second, so the answer is: $$ \frac 4 6 114 = 76$$.