JEE Mathematics 2023 Question with Solution

Given: The $k$-th A.P. has first term $k$ and common difference $2k-1$. We need the sum of the first $12$ terms for each such A.P. and then evaluate $\sum_{k=1}^{10} s_k$.

Find: The value of $\sum_{k=1}^{10} s_k$.

For an A.P. with first term $a$ and common difference $d$, the sum of the first $12$ terms is

$$S_{12} = \frac{12}{2}\left[2a + (12-1)d\right] = 6(2a+11d)$$

For the $k$-th progression,

$$s_k = 6\left(2k + 11(2k-1)\right)$$ $$= 6(2k+22k-11)$$ $$= 6(24k-11)$$ $$= 144k-66$$

Now,

$$\sum_{k=1}^{10} s_k = \sum_{k=1}^{10} (144k-66)$$ $$= 144\sum_{k=1}^{10} k - 66\sum_{k=1}^{10} 1$$

Using

$$\sum_{k=1}^{10} k = \frac{10\cdot 11}{2} = 55$$

and

$$\sum_{k=1}^{10} 1 = 10$$

we get

$$\sum_{k=1}^{10} s_k = 144\cdot 55 - 66\cdot 10$$ $$= 7920 - 660$$ $$= 7260$$

Therefore, the correct option is A, and $\sum_{i=1}^{10} s_i = 7260$.

Let the A.P.s be indexed by $k=1,2,3,\dots,10$. Then:

• first term of the $k$-th A.P. is $k$
• common difference of the $k$-th A.P. is the $k$-th odd number, that is $2k-1$

So the required sum to $12$ terms is

$$s_k = \frac{12}{2}\left[2k + 11(2k-1)\right]$$ $$= 6(24k-11)$$ $$= 144k-66$$

Hence the total becomes

$$\sum_{k=1}^{10} s_k = \sum_{k=1}^{10}(144k-66)$$

Split the summation termwise:

$$= 144\sum_{k=1}^{10}k - 66\sum_{k=1}^{10}1$$

Now substitute the standard sums:

$$= 144(55) - 66(10)$$ $$= 7260$$

Thus the required value is $7260$.