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  • The 10th term of the series 2 + 8 + 20+ 40 +70 +112+ ……. Option: 1 240Option: 2</str

The 10th term of the series 2 + 8 + 20+ 40 +70 +112+ ……. 

Option: 1

240


Option: 2

340


Option: 3

440


Option: 4

640


Answers (1)

best_answer

\begin{array}{l}{\text { Solution }} \\ {\text { The sequence of first consecutive differences is } 6,12,20,30,42 \text { . }} \\ {\text { The sequence of second consecutive differences is } 6,8,10,12 \ldots \ldots \text { it is an AP }} \\ {S_{n}=2+8+20+40+70+112+\ldots \ldots+T_{n-1}+T_{n}} \\ {S_{n}=\quad 2+8+20+40+70+112+\ldots \ldots+T_{n-1}+T_{n}} \\ {0=2+6+12+20+30+42+\ldots \ldots \ldots+\left(T_{n-1}-T_{n}\right)-T_{n}} \\ {T_{n}=2+6+12+20+30+42+\ldots \ldots \ldots+\text { upto } n \text { term }} \\ {\text { Or }}\end{array}\\\begin{array}{l}{T_{n}=2+6+12+20+30+42+\ldots \ldots \ldots+t_{n-1}+t_{n}} \\ {T_{n}=2+6+12+20+30+42+\ldots \ldots \ldots+t_{n-1}+t_{n}} \\ {\text { Substract (iii) }-(\text { iv) }} \\ {0=2+4+6+8+10+12 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\left(t_{n-1}-t_{n}\right)-t_{n}} \\ {t_{n}=2+4+6+8+10+12+\ldots \ldots \ldots \ldots\left(t_{n-1}-t_{n}\right)} \\ {t_{n}=\frac{n}{2}(2 \times 2+(n-1) \times 2)=n^{2}+n}\end{array}\begin{array}{l}{T_{n}=\sum t_{n}=\Sigma\left(n^{2}+n\right)=\left(\frac{n(n+1)(2 n+1)}{6}\right)+\left(\frac{n(n+1)}{2}\right)} \\ {T_{n}=\frac{n(n+1)(n+2)}{3}} \\ {\text { Put } n=10} \\ {T_{n}=440} \\ {\text { Correct option 3} }\end{array}

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SANGALDEEP SINGH

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