Manipal Entrance Test 2013 Math Paper

Clearly, the given series is an arithmetico-geometric series whose corresponding AP and GP are respectively.
1 ,4 ,7 ,1 0 , and 1,

1

5

,

1

52

,

1

53

,....
The nth term of A P=[1+(n−1)×3]=3n−2
The nth term of G P =[1×(

1

5

)n−1]=(

1

5

)n−1
So the nth term of the given series is
(3n−2)×

1

5n−1

=

3n−2

5n−1

Let, Sn=1+

4

5

+

7

52

+

10

53

+....+

3n−5

5n−2

+

3n−2

5n−1

......(i)

1

5

Sn=

1

5

+

4

52

+

7

53

+....+

(3n−5)

5n−1

+

3n−2

5n

....(ii)
Subtracting (ii) from (i), we get
Sn−

1

5

Sn=1+{

3

5

+

3

52

+

3

53

+....+

3

5n−1

}−

(3n−2)

5n

⇒

4

5

Sn=1+

3

5

{1−( 1 5 )n−1}

(1− 1 5 )

−

(3n−2)

5n

⇒

4

5

Sn=1+

3

5

{1− 1 5n−1 }

( 4 5 )

−

(3n−2)

5n

⇒

4

5

Sn=1+

3

4

(1−

1

5n−1

)−

(3n−2)

5n

⇒ Sn=

5

4

+

15

16

(1−

1

5n−1

)−

(3n−2)

4.5n−1