# Q. Given `{a_1,a_2,a_3}` is a G.P. and `{a_2,a_3,a_4}` is an A.P. `a_1 + a_4 = 14` and `a_2 + a_3 = 12.` Then the possible values of `a_4` is given by:- A) 1.5 B) 2.5 C) 12 D) 12.5

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### 1 Answer

We are given that `{a_1,a_2,a_3}` is in geometric progression while `{a_2,a_3,a_4}` is in arithmetic progression. Also `a_1+a_4=14` and `a_2+a_3=12` . We are asked to find the possible values for `a_4` :

We can write the geometric progression as `{a_1,a_2=ra_1,a_3=r^2a_1}`

and the arithmetic progression as `{a_2,a_3=a_2+d,a_4=a_2+2d}`

Now `a_1+a_4=14`

`a_1+a_2+2d=14` Substituting for `a_4`

`a_1+ra_1+2d=14` Substituting for `a_2` (1)

``Now `a_3-a_2=a_3-a_2`

`a_2+d-a_2=r^2a_1-ra_1` Using the AP and GP substitutions

`d=ra_1(r-1)` Substitute this into (1)

`a_1+ra_1+2ra_1(r-1)=14`

`a_1+ra_1+2r^2a_1-2ra_1=14` (2)

``Also `a_2+a_3=12`

`ra_1+r^2a_1=12` (a)

** Then `a_1(r+r^2)=12` and `a_1=12/(r+r^2)` **Rewrite (2) using (a):

`a_1+(ra_1+r^2a_1)+r^2a_1-2ra_1=14`

`a_1+12+r^2a_1-2ra_1=14`

`a_1(r^2-2r+1)=2`

So `a_1=2/(r^2-2r+1)` and `a_1=12/(r^2+r)`

`2/(r^2-2r+1)=12/(r^2+r)`

`2r^2+2r=12r^2-24r+12`

`10r^2-26r+12=0`

`2(5r-3)(r-2)=0`

`r=2` : `a_2+a_3=ra_1+r^2a_1=2a_1+4a_1=6a_1=12`

so r=2 ==> `a_1=2` and we have `a_1=2,a_2=4,a_3=8,a_4=12`

`r=3/5` : `(3/5)a_1+(9/25)a_1=12`

`24a_1=300==>a_1=25/2`

So `r=3/5 ==> a_1=25/2,a_2=15/2,a_3=9/2,a_4=3/2`

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The two possible values for `a_4` are 12 and `3/2` so A and C are both correct.

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