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If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is :

A. 13

B. 9

C. 21

D. 17

Answer: Option C

Solution(By Examveda Team)

Let three consecutive terms of an increasing A.P. be a - d, a + d where a is the first term and d be the common difference
$$\eqalign{ & \therefore a - d + a + a + d = 51 \cr & \Rightarrow 3a + 51 \cr & \therefore a = \frac{{51}}{3} = 17 \cr} $$
and product of the first and third terms
$$\eqalign{ & = \left( {a - d} \right)\left( {a + d} \right) = 273 \cr & \Rightarrow {a^2} - {d^2} = 273 \cr & \Rightarrow {\left( {17} \right)^2} - {d^2} = 273 \cr & \Rightarrow 289 - {d^2} = 273 \cr & \Rightarrow {d^2} = 289 - 273 \cr & \Rightarrow {d^2} = 16 \cr & \Rightarrow {d^2} = {\left( { \pm 4} \right)^2} \cr & \therefore d = \pm 4 \cr & \because {\text{The A}}{\text{.P}}{\text{.}}\,{\text{is}}\,{\text{increasing}} \cr & \therefore d = 4 \cr & {\text{Now}}\,{\text{third}}\,{\text{term}} = a + d \cr & = 17 + 4 = 21 \cr} $$

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