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  • Let <img alt="\mathrm{s}_1, \mathrm{~s}_2, \mathrm{~s}_3, \ldots \ldots, \mathrm{s}_{10}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bs%7D_1%2C%20%5Cmathrm%7B%7Es%7D_2%2C%20%5C

Let \mathrm{s}_1, \mathrm{~s}_2, \mathrm{~s}_3, \ldots \ldots, \mathrm{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 \ldots, 19 respectively. Then \sum_{\mathrm{i}=1}^{10} \mathrm{~s}_{\mathrm{i}} is equal to


 

Option: 1

7260
 


Option: 2

7380
 


Option: 3

7220
 


Option: 4

7360


Answers (1)

best_answer

\begin{aligned} & \mathrm{S}_{\mathrm{k}}=6(2 \mathrm{k}+(11)(2 \mathrm{k}-1)) \\ & \mathrm{S}_{\mathrm{k}}=6(2 \mathrm{k}+22 \mathrm{k}-11) \\ & \mathrm{S}_{\mathrm{k}}=144 \mathrm{k}-66 \\ \end{aligned}

\begin{aligned} & \sum_1^{10} \mathrm{~S}_{\mathrm{k}}=144 \sum_{\mathrm{k}=1}^{10} \mathrm{k}-66 \times 10 \\ & =144 \times \frac{10 \times 11}{2}-660 \\ & =7920-660 \\ & =7260 \end{aligned}

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