Q43 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics

$2+3+5+6+8+9+\ldots . .+2 n \text { terms }=$

  • Correct answer
    A
    $3 n^2+2 n$
  • B
    $4 n^2+2 n$
  • C
    $4 n^2$
  • D
    $5 n^2+2 n$

Solution

$\begin{aligned} & \text { Required sum }=1+2+3+4 \ldots \ldots . \text { upto }(3 n+1) \text { terms } \\ & \quad-(4+7+10 \ldots . n \text { terms })-1 \\ & =\frac{(3 n+1)(3 n+2)}{2}-\frac{n}{2}[8+(n-1) \times 3]-1 \\ & =\frac{9 n^2+9 n+2-3 n^2-5 n-2}{2}=3 n^2+2 n\end{aligned}$

Hence, the correct answer is option: 'A'.

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