A.P and G.P

Adding the corresponding terms of an A.P and G.P makes the new sequence 1,7,6,9,... .
Find the $ \displaystyle {{n}^{{th}}}$ term of this sequence .

Solution

Let $ \displaystyle a,a+d,a+2d,a+3d,...\ \text{is an }A.P\text{  and}$

$ \displaystyle A,Ar,A{{r}^{2}},A{{r}^{3}},...\ \text{is a }G.P\ .$

$ \displaystyle a+A,a+d+Ar,a+2d+A{{r}^{2}},a+3d+A{{r}^{3}},...=1,7,6,9,...$

$ \displaystyle a+A=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......(1)$

$\displaystyle a+d+Ar=7\ \ \ \ \ \ \ ......(2)$

$ \displaystyle a+2d+A{{r}^{2}}=6\ \ \ \ \ .....(3)$

$ \displaystyle a+3d+A{{r}^{3}}=9\ \ \ \ \ .....(4)$

$ \displaystyle (2)-(1)\Rightarrow \ \ \ d+Ar-A=6\ \ \ \ \ \ \ .......(5)$

$ \displaystyle (3)-(2)\Rightarrow \ \ \ d+A{{r}^{2}}-Ar=-1\ ........(6)$

$ \displaystyle (4)-(3)\Rightarrow \ \ \ d+A{{r}^{3}}-A{{r}^{2}}=3\ \ ........(7)$

$ \displaystyle (6)-(5)\Rightarrow \ \ \ A{{r}^{2}}-2Ar+A=-7$

$ \displaystyle (7)-(6)\Rightarrow \ \ \ A{{r}^{3}}-2A{{r}^{2}}+Ar=4$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r(A{{r}^{2}}-2Ar+A)=4$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -7r=4$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r=-\frac{4}{7}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{16}}{{49}}A+\frac{8}{7}A+A=-7$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 16A+56A+49A=-343$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A=-\frac{{343}}{{121}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a=1+\frac{{343}}{{121}}=\frac{{464}}{{121}}\ \ \ \ \ \left[ {By(1)} \right]$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d+\frac{{196}}{{121}}+\frac{{343}}{{121}}=6\ \ \ \ \ \ \ \left[ {By(5)} \right]$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d=\frac{{187}}{{121}}$

$ \displaystyle \text{The }{{\text{n}}^{{th}}}\text{ term of given sequence = }{{u}_{n}}$

$ \displaystyle \ \ \ {{u}_{n}}=a+(n-1)d+A{{r}^{{n-1}}}$

$ \displaystyle \ \ \ \ \ \ =\frac{{464}}{{121}}+(n-1)\frac{{187}}{{121}}+(-\frac{{343}}{{121}}){{(-\frac{4}{7})}^{{n-1}}}$

$ \displaystyle \ \ \ \ \ \ =\frac{1}{{121}}\left[ {277+187n-343{{{(-\frac{4}{7})}}^{{n-1}}}} \right]$