Given a sequence of 4 numbers, first three of which are in G.P. and last three are in A.P. with common difference three. If first and last terms of this sequence are equal, then last term is

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Solution

The correct option is B 4
Let $a_{1}, a_{2}, a_{3}, a_{4}$ be given 4 numbers
$a_{1}, a_{2}, a_{3}$ form G.P $\Rightarrow \frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = t (s a y) \Rightarrow a_{2} = t a_{1}, a_{3} = t^{2} a_{1}$
$\Rightarrow 2 a_{3} = a_{2} + a_{4} \Rightarrow 2 t^{2} a_{1} = t a_{1} + a_{1}$
$\Rightarrow 2 t^{2} - t - 1 = 0$
$\Rightarrow t = \frac{- 1}{2}$
$∴$ $a_{4} = a_{2} + 6$ $\Rightarrow a_{1} = \frac{- 1}{2} a_{1} + 6$
$\Rightarrow$ $\frac{3}{2} a_{1} = 6$
$\Rightarrow$ $a_{1} = 4 = a_{4}$
$∴$ $a_{4} = 4$

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Given a sequence of 4 numbers, first three of which are in G.P. and last three are in A.P. with common difference three. If first and last terms of this sequence are equal, then last term is

The correct option is B 4Let a1,a2,a3,a4 be given 4 numbers a1,a2,a3 form G.P ⇒a2a1=a3a2=t(say)⇒a2=ta1,a3=t2a1 ⇒2a3=a2+a4⇒2t2a1=ta1+a1 ⇒2t2–t–1=0 ⇒t=−12 ∴ a4=a2+6 ⇒a1=−12a1+6 ⇒ 32a1=6 ⇒ a1=4=a4 ∴ a4=4