Question

The sum of first ten terms of an A.P. is equal to 155, and the sum of the first two terms of a G.P. is 9, find these progressions if the first term of A.P. is equal to common ratio of G.P. and the first term of G.P. is equal to common difference of A.P

• A. $r=\frac{25}{2}$,A.P is $\frac{25}{2},\frac{79}{6},\frac{83}{6},.....$ and G.P is $\frac{2}{3},\frac{25}{3},\frac{625}{6},......$
• B. $r=3$,A.P is $2,5,,8,11,...$ and G.P is $3,6,12,24,....$
• C. Both the above
• D. none of these

Answer 1

Answer: (C)

Both the above

Let the A.P and the G.P be respectively
$a,a+d,a+2d,...$ and $d,da,{da}^2,...$
Now $S_{10}$ of A.P $=\frac{10}{2}[2a+9d]$
$=10a+45d=155$ ...(1)
And $S_2$ of G.P $=d+da=9$ ...(2)
(1) $\Rightarrow a=\frac{31-9d}{2}$
Putting this value of a in (2) we get
$d+d\left(\frac{31-9d}{2}\right)=9\Rightarrow 2d+31d-{9d}^2=18\Rightarrow {3d}^2-11d+6=0$
$\therefore d=\frac{2}{3},3$
Case 1: $d=\frac{2}{3}$
$\therefore a=\frac{31-9\left(\frac{2}{3}\right)}{2}=\frac{25}{2}$
$A.P\frac{25}{2},\frac{25}{2}+\frac{2}{3},\frac{25}{2}+\frac{4}{3}...$ i.e. $\frac{25}{2},\frac{79}{6},\frac{83}{6},...$
and $G.P:\frac{2}{3},\frac{2}{3}\left(\frac{25}{2}\right),\frac{2}{3}{\left(\frac{25}{3}\right)}^2,...$ i.e. $\frac{2}{3}\left(\frac{25}{3}\right),\frac{625}{6},...$
Case 2: $d=3$
$\therefore a=\frac{31-9\left(3\right)}{2}=2$
$\therefore A.P:2,2+3,2+2\left(3\right),...$ i.e. $2,5,8,...$
and $G.P:3,3\left(2\right),3{\left(2\right)}^2,...$ i.e $3,6,12,...$