JEE Main 2023 — Sequences And Series Question with Solution

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

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 $$i=1,2,\dots ,10$$:

$$\sum _{i=1}^{10}{s}_i=6\sum _{i=1}^{10}(24i-11)=6\left(24\sum _{i=1}^{10}i-11\sum _{i=1}^{10}1\right)$$

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

$$\sum _{i=1}^{10}{s}_i=6\left(24\cdot 55-11\cdot 10\right)=6(1320-110)=6\cdot 1210=7260$$

Thus, the sum $$\sum _{i=1}^{10}{s}_i$$ is equal to 7260.