Question

If 12,x2,x3,33...are in AP then x2 is?

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{12,x_2,x_3,33\;are\;in\;A.P}$

$\underline{\textsf{To find:}}$

$\mathsf{The\;value\;of\;x_2}$

$\underline{\textsf{Solution:}}$

$\textsf{Let 'd' be the common difference of the given A.P}$

$\mathsf{Then,}$

$\mathsf{x_2=12+d}$

$\mathsf{x_3=12+2d}$

$\mathsf{33=12+3d}$

$\implies\mathsf{33-12=3d}$

$\implies\mathsf{3d=21}$

$\implies\mathsf{d=\dfrac{21}{7}}$

$\implies\mathsf{d=3}$

$\mathsf{x_2=12+d}$

$\mathsf{x_2=12+7}$

$\implies\boxed{\boxed{\mathsf{x_2=19}}}$

$\underline{\textsf{Find more:}}$

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