Let a,b,c be positive integers such that b/a is an integer. If a,b,c are in geometric progression and the arithmetic mean of a,b,c is b + 2, then the value of fraca^2 + a - 14a + 1 is

\ (b / a)=(c / b)= integer (say) (a+b+c / 3)=b+2 (Given) => a+b+c=3 b+6 => a-2 b+c=6 => a-2 b+(b^2 / a)=6 => (b^2 / a^2)-(2 b / a)+1=(6 / a) Therefore ((b / a)-1)^2=(6 / a) = perfect square. Therefore a=6 only \ Thus, (a^2+a-14 / a+1)=(36+6-14 / 7)=(28 / 7) \ =4 Let a, b, c be a, a r, a r^2 Where r in N Now, a+a r+a r^2=3 a r+6 \ => a r^2-2 a r+a=6 => a(r-1)^2=6 => (r-1)^2=6 / a Therefore a=6 only.

Let a,b,c be positive integers such that b/a is an inte... | Sequence & Series (Mathematics)

$\begin{aligned}& \frac{b}{a}=\frac{c}{b}=\text{ integer (say) } \\& \frac{a+b+c}{3}=b+2 \text { (Given) } \\& \Rightarrow a+b+c=3 b+6 \Rightarrow a-2 b+c=6 \\& \Rightarrow a-2 b+\frac{b^2}{a}=6 \\& \Rightarrow \frac{b^2}{a^2}-\frac{2 b}{a}+1=\frac{6}{a} \\& \therefore\left(\frac{b}{a}-1\right)^2=\frac{6}{a} =\text{ perfect square. } \\& \therefore a=6 \text { only }\end{aligned}\\$ Thus, $\frac{a^2+a-14}{a+1}=\frac{36+6-14}{7}=\frac{28}{7} \\ =4$ Let $a, b, c$ be $a, a r, a r^2$ $\\$ Where $r \in N$ Now, $a+a r+a r^2=3 a r+6$ $\begin{aligned}& \Rightarrow a r^2-2 a r+a=6 \Rightarrow a(r-1)^2=6 \\& \Rightarrow(r-1)^2=6 / a \\& \therefore a=6 \text{ only. }\end{aligned}$

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}( x_{i} eq 0 .\ for . \ i = 1,2,,n ) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then \sum_{i = 1}^{n}\mspace{2mu}( \frac{1}{x_{i}} ) is equal to

Let 3,a,b,c be in A.P. and 3,a - 1,b + 1,c + 9 be in G.P. Then, the arithmetic mean of a,b and c is

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