The nth term of the sequence

^{3}/_{2}, 3, 7, 16, 35, 74 ….. is-
**A.**

5 . 2^{n-2}– n -
**B.**

5 . 2^{n-2}– (n+1) / 2 -
**C.**

3 . 2^{n-2} -
**D.**^{3}/_{2}n

##### Correct Answer: Option A

##### Explanation

^{3}/_{2}, 3, 7, 16, 35, 74, ….

Using the method of substitution

When n = 1, 5 . 2^{n-2} – 1 = 5 . 2^{1-2} – 1

= 5 x 2^{-1} – 1

= 5 x ^{1}/_{2} – 1

= ^{5}/_{2} – 1 = ^{3}/_{2}

When n = 1, 5 . 2^{n-2} – n

= 5 . 2^{2-2} – 2

= 5 x 2^{0} – 2

= 5 x 1 – 2 = 3

When n = 3, 5 . 2^{n-2} – n

= 5 . 2^{3-2} – 3

= 5 x 2^{1} – 3

= 5 x 2 – 3

= 10 -3 = 7

When n = 4, 5 . 2^{n-2} – n

= 5 . 2^{4-2} – 4

= 5 x 2^{2} – 4

= 5 x 4 – 4

= 20 – 4 = 16

When n = 5, 5 . 2^{n-2} – n

= 5 . 2^{5-2} – 5

= 5 x 2^{3} – 5

= 5 x 8 – 5

= 40 – 5

= 35

When n = 6, 5 . 2^{n-2} – n

= 5 . 2^{6-2} – 6

= 5 x 2^{4} – 6

= 5 x 16 – 6

= 80 – 6 = 74

∴ 5 . 2^{n-2} – 1 = ^{3}/_{2}, 3, 7, 16, 35, 74