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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. 17Answer: Option C
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Solution(By Apex 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 $\begin{aligned}&\therefore a-d+a+a+d=51\\ &\Rightarrow3a+51\\ &\therefore a=\frac{51}{3}=17\end{aligned}$ and product of the first and third terms $\begin{array}{l}=(a-d)(a+d)=273\\ \Rightarrow a^2-d^2=273\\ \Rightarrow(17)^2-d^2=273\\ \Rightarrow289-d^2=273\\ \Rightarrow d^2=289-273\\ \Rightarrow d^2=16\\ \Rightarrow d^2=(\pm4)^2\\ \therefore d=\pm4\\ \because\text{ The A.P. is increasing }\\ \therefore d=4\\ \text{ Now third term }=a+d\\ =17+4=21\end{array}$

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