JEE Mathematics 2023 Question with Solution

Given:

• $a = A + 6d$, $b = A + 1 + 8d$, $c = A + 2 + 16d$
• $a = 29$

Find: The sum of the first $20$ terms of the AP with first term $c-a-b$ and common difference $\frac{d}{12}$.

From the given A.P.s,

Using $a = 29$,

The solution working gives

Now compute the first term of the required AP:

Substituting $A=-7$ and $d=6$,

Its common difference is

Therefore, for the required AP,

Use the sum formula:

So,

Therefore, the required sum is $495$.

Given:

and

with $a=29$.

Find: $S_{20}$ for the AP with first term $c-a-b$ and common difference $\frac{d}{12}$.

Substitute the expressions for $a, b, c$ into the determinant:

The extracted solution states that solving this gives

Also, this is consistent with

Now,

Hence,

The new AP has

Thus,

So the answer is $495$.

Note: The alternate provided approach contains inconsistent intermediate values, but the final answer and the primary solution conclusion support $495$.