if $a_1,a_2...a_n $ are in AP and the first 16 terms add up to 114 , find $a_1+a_6+a_{11}+a_{16}$

Question

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

Answer 1

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: $$ \frac 4 6 114 = 76$$.

Answer 2

Your answer is correct. You can easily see that the official answer is wrong, by considering the constant sequence $\frac{114}{16},\frac{114}{16},...,\frac{114}{16}$. In this case, clearly the sum of any four terms is $114/4=28.5$, not $76$.

Answer 3

As commented by Macavity, the original problem has been mis-stated.

Given:

• $~a_1, a_2, \cdots, a_n~$ is a sequence in arithmetic progression.

• $~a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 114.$

To Do:

Compute $~a_1 + a_6 + a_{11} + a_{16}.$

---

Since the sequence is in arithmetic progression,
$~a_n = a_1 + (n-1)d.$

Therefore,

$$114 = 6a_1 + d(0 + 3 + 6 + 9 + 12 + 15) = 6a_1 + 45d. \tag1 $$

Then, the expression to be computed is

$$4a_1 + d(0 + 5 + 10 + 15) = 4a_1 + 30d. \tag2 $$

Regardless of the specific values of $~a~$ and $~d,~$ the computation in (2) above is exactly $~\dfrac{2}{3} ~$ times the computation in (1) above.

Therefore, the computation in (2) above equals

$$\frac{2}{3} \times 114 = 76.$$