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  • Let an be the nth term of the series 5 + 8 + 14 + 23 + 35 + 50 +.... and <img alt="\mathrm{Sn}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{K}} \text {. Then } \mathrm{S}_{30}-\mathrm{

Let an be the nth term of the series 5 + 8 + 14 + 23 + 35 + 50 +.... and \mathrm{Sn}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{K}} \text {. Then } \mathrm{S}_{30}-\mathrm{a}_{40} is equal to

Option: 1

11260 


Option: 2

11280


Option: 3

11290


Option: 4

11310


Answers (1)

best_answer

\begin{aligned} &\begin{aligned} & \mathrm{S}_{\mathrm{n}}=5+8+14+23+35+50+\ldots+\mathrm{a}_{\mathrm{n}} \\ & \mathrm{S}_{\mathrm{n}}=5+8+14+23+35+\ldots+\mathrm{a}_{\mathrm{n}} \end{aligned}\\ &\begin{aligned} & \mathrm{O}=5+3+6+9+12+15+\ldots-\mathrm{a}_{\mathrm{n}} \\ & \mathrm{a}_{\mathrm{n}}=5+(3+6+9+\ldots(\mathrm{n}-1) \text { terms }) \\ & \mathrm{a}_{\mathrm{n}}=\frac{3 \mathrm{n}^2-3 \mathrm{n}+10}{2} \\ & \mathrm{a}_{40}=\frac{3(40)^2-3(40)+10}{2}=2345 \\ & \mathrm{~S}_{30}=\frac{3 \sum_{\mathrm{n}=1}^{30} \mathrm{n}^2-3 \sum_{\mathrm{n}=1}^{30} \mathrm{n}+10 \sum_{\mathrm{n}=1}^{30} 1}{2} \\ & =\frac{\frac{3 \times 30 \times 31 \times 61}{6}-\frac{3 \times 30 \times 31}{2}+10 \times 30}{2} \\ & \mathrm{~S}_{30}=13635 \\ & \mathrm{~S}_{30}-\mathrm{a}_{40}=13635-2345 \\ & =11290(\text { Option }(3)) \\ & \end{aligned} \end{aligned}

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