An-exponential-sequence-of-positive-terms-and-a-linear-sequence-have-the-same-first-term-the-sum-o-their-first-term-is-3-the-sum-of-their-second-term-is-3-2-and-the-sum-of-their-third-term-is-6-

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Question Number 10488 by Saham last updated on 13/Feb/17

$$\mathrm{An}\mathrm{exponential}\mathrm{sequence}\mathrm{of}\mathrm{positive}\mathrm{terms}\mathrm{and}\mathrm{a}$$$$\mathrm{linear}\mathrm{sequence}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{first}\mathrm{term}.\mathrm{the}\mathrm{sum}$$$$\mathrm{o}\mathrm{their}\mathrm{first}\mathrm{term}\mathrm{is}3,\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{their}\mathrm{second}\mathrm{term}$$$$\mathrm{is}\frac{3}{2},\mathrm{and}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{their}\mathrm{third}\mathrm{term}\mathrm{is}6.\mathrm{find}\mathrm{the}$$$$\mathrm{sum}\mathrm{of}\mathrm{their}\mathrm{fifth}\mathrm{term}.$$

Answered by mrW1 last updated on 14/Feb/17

$$termsofA.P.:$$$$a_1=a$$$$a_2=a+d$$$$a_3=a+2d$$$$a_4=a+3d$$$$a_5=a+4d$$$$\cdot \cdot \cdot$$$$ternsofG.P.:$$$$b_1=a$$$$b_2=a\times q$$$$b_3=a\times q^2$$$$b_4=a\times q^3$$$$b_5=a\times q^4$$$$\cdot \cdot \cdot$$$$a_1+b_1=a+a=3$$$$\Rightarrow a=\frac{3}{2}$$$$a_2+b_2=a+d+a\times q=\frac{3}{2}$$$$\Rightarrow a+d+aq=\frac{3}{2}$$$$\Rightarrow d+\frac{3}{2}q=0\dots \left(i\right)$$$$a_3+b_3=a+2d+a\times q^2=6$$$$\Rightarrow d+\frac{3}{4}{q}^2=\frac{9}{4}\dots \left(ii\right)$$$$\left(ii\right)-\left(i\right):$$$$\frac{3}{4}{q}^2-\frac{3}{2}q-\frac{9}{4}=0$$$$q^2-2q-3=0$$$$(q-3)(q+1)=0$$$$\Rightarrow q=3,sincetermsofG.P.arepositive$$$$d=-\frac{3}{2}\times 3=-\frac{9}{2}$$$$a_5=a+4d=\frac{3}{2}+4\times (-\frac{9}{2})=-\frac{33}{2}$$$$b_5=a\times q^4=\frac{3}{2}\times 3^4=\frac{243}{2}$$$$a_5+b_5=-\frac{33}{2}+\frac{243}{2}=\frac{210}{2}=105$$

Commented by Saham last updated on 14/Feb/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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