Chapter 9: Sequence and Series

Sum of first n terms be S_n given as S_n = 2n + 3n²

We have to find the r^th term that is a_r

Using the given hint n^th term is given as a_n = S_n – S_n-1

⇒ a_r = S_r – S_r-1

Using S_n = 2n + 3n²

⇒ a_r = 2r + 3r² – (2(r – 1) + 3(r – 1)²)

⇒ a_r = 2r + 3r² – (2r – 2 + 3(r² – 2r + 1))

⇒ a_r = 2r + 3r² – (2r – 2 + 3r² – 6r + 3) ⇒ a_r = 6r – 1

Hence the r^th term is 6r – 1

Long Answer Type

Therefore

⇒ RHS = 2A

Hence RHS = LHS

Hence proved

Objective Type Questions

(D) 4

Explanation:

To find: Common Difference of A.P that is d

Consider,

S_n = 3n + 2n²

Putting n = 1, we get

S₁ = 3(1) + 2(1)²

= 3 + 2

S₁ = 5

Putting n = 2, we get

S₂ = 3(2) + 2(2)²

= 6 + 2(4)

= 6 + 8

S₂ = 14

Now, we know that,

S₁ = a₁

⇒ a₁ = 5

And a₂ = S₂ – S₁

= 14 – 5

= 9

∴ Common Difference, d = a₂ – a₁

= 9 – 5

= 4

Hence, the correct option is (D)

(C) 4⁵

Explanation:

Given the third term of G.P, T₃ = 4

To find the product of first five terms

We know that,

T_n = a r^{n – 1}

It is given that, T₃ = 4

⇒ ar^{3 – 1} = 4

⇒ ar² = 4 … (i)

Product of first 5 terms = a × ar × ar² × ar³ × ar⁴

= a⁵r^1+2+3+4

= a⁵r¹⁰

= (ar²)⁵

= (4)⁵ [from (i)]

Hence, the correct option is (C)