QJEEM23Ma182f: JEE Main 2023 Mathematics - Class 11 JEEM23Ma182f Solution: Let respectively be the sum to 12 terms of 10 A.P. s whose…

We have 10 arithmetic progressions (A.P.s) with the first terms $a_i$ and the common differences $d_i$ , where $i = 1, 2, \ldots, 10$ .

The first terms are $a_i = i$ and the common differences are $d_i = 2i - 1$ .

Now, we need to find the sum of the first 12 terms for each A.P. The formula for the sum of the first n terms of an A.P. is:

$S_n = n\left(\frac{2a + (n - 1)d}{2}\right)$

In this case, we need to find the sum of the first 12 terms for each A.P., so we have:

$S_{12} = 12\left(\frac{2a + 11d}{2}\right)$

Now, we can compute the sum $s_i$ for each A.P.:

$s_i = 12\left(\frac{2i + 11(2i - 1)}{2}\right) = 6(2i + 22i - 11) = 6(24i - 11)$

Finally, we need to find the sum of all $s_i$ for $i = 1, 2, \ldots, 10$ :

$\sum\limits_{i=1}^{10} s_i = 6\sum\limits_{i=1}^{10} (24i - 11) = 6\left(24\sum\limits_{i=1}^{10} i - 11\sum\limits_{i=1}^{10} 1\right)$

The sum of the first 10 integers is $\sum\limits_{i=1}^{10} i = \frac{10(10 + 1)}{2} = 55$ , so we have:

$\sum\limits_{i=1}^{10} s_i = 6\left(24\cdot55 - 11\cdot10\right) = 6(1320 - 110) = 6\cdot1210 = 7260$

Thus, the sum $\sum\limits_{i=1}^{10} s_i$ is equal to 7260.