Manipal Entrance Test 2020 Math Paper

Let four terms in a GP be ar3,ar,‌

a

r

and ‌

a

r3

.
According to the given condition,
ar3+ar+‌

a

r

+‌

a

r3

=60......(i)
and ‌‌‌

ar3+ a r3

2

=18
⇒‌‌ar3+‌

a

r3

=36......(ii)
Now, from Eq. (i), we have
(ar+‌

a

r

)+ar3+‌

a

r3

=60
⇒‌‌a(r+‌

1

r

)+36=60 [from Eq.(ii)]
⇒‌‌a(r+‌

1

r

)=24........(iii)
On dividing Eq. (iii) by Eq. (ii), we get
‌

a(r3+ 1 r3 )

a(r+‌ 1 r )

=‌

36

24

⇒‌

(r+ 1 r )(r2+ 1 r2 −1)

r+‌ 1 r

=‌

3

2

⇒‌‌2(r2+‌

1

r2

−1)=3
⇒‌‌‌

2(r4+1−r2)

r2

=3
⇒‌‌2r4+2−2r2=3r2
⇒‌‌2r4−5r2+2=0
⇒‌‌2r4−4r2−r2+2=0
⇒2r2(r2−2)−1(r2−2)=0
⇒‌‌(r2−2)(2r2−1)=0
⇒‌‌r2=2,2r2=1
⇒‌‌r=±√2,r=±‌

1

√2

On putting r=√2 in Eq. (iii), we get a(√2+‌

1

√2

)=24
⇒ ‌‌a(‌

2+1

√2

)=24
⇒ ‌‌3a=24√2
⇒‌‌a=8√2
∴ Series becomes 8√2(√2)3,8√2(√2),‌

8√2

√2

and
‌

8√2

(√2)3

i.e., 32, 16,8 and 4
If we take r=‌

1

√2

, we get the series
4,8,16 and 32