Let \[s_{1}, s_{2}, s_{3}, \ldots, s_{10}\] respectively be the sum to 12 terms of 10 A.P. s whose first terms are \[1,2,3, \ldots .10\] and the common differences are \[1,3,5, \ldots \ldots, 19\]
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Official solution
We have 10 arithmetic progressions (A.P.s) with the first terms ai and the common differences di, where \[i = 1, 2, \ldots, 10\].
The first terms are ai=i and the common differences are di=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 si 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 si 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.