Question

Write first 4 terms in each of the sequences:

(i) a_n = (5n + 2)

(ii) a_n = (2n - 3)/4

(iii) a_n = (–1)^n–1 × 2^{n + 1}

Answer 1

To Find: First four terms of given series.

(i) Given: n^th term of series is (5n + 2)

Put n = 1, 2, 3, 4 in n^th term, we get

first (a₁), Second (a₂), Third (a₃) & Fourth (a₄) term

a₁ = (5 × 1 + 2) = 7

a₂ = (5 × 2 + 2) = 12

a₃ = (5 × 3 + 2) = 17

a₄ = (5 × 4 + 2) = 22

First four terms of given series is 7, 12,17,22

ALTER: When you find or you have first term (a or a₁) and second term (a₂) then find the difference (a₂ - a₁)

Now add this difference in last term to get the next term

For example a₁ = 7 and a₂ = 12, so difference is 12 – 5 = 7

Now a₃ = 12 + 5 = 17, a₄ = 17 + 5 = 22

(This method is only for A.P)

NOTE: When you have nth term in the form of (a × n + b)

Then common difference of this series is equal to a.

This type of series is called A.P (Arithmetic Progression)

(Where a, b are constant, and n is number of terms)

(ii) Given: n^th term of series is (2n - 3)/4

Put n = 1, 2, 3, 4 in nth term, we get

first (a₁), Second (a₂), Third (a₃) & Fourth (a₄) term.

First four terms of given series are -1/4, 1/4, 3,4, 5/4

(iii) Given: n^th term of series is (–1)^n–1 × 2^{n + 1}

Put n = 1, 2, 3, 4 in nth term, we get

first (a₁), Second (a₂), Third (a₃) & Fourth (a₄) term.

a₁ = (–1)^1–1 × 2^{1 + 1} = (–1)⁰ × 2² = 1 × 4 = 4

a₂ = (–1)^2–1 × 2^{2 + 1} = (–1)¹ × 2³ = (–1) × 8 = (–8)

a₃ = (–1)^3–1 × 2^{3 + 1} = (–1)² × 2⁴ = 1 × 16 = 16

a₄ = (–1)^4–1 × 2^{4 + 1} = (–1)³ × 2⁵ = (–1) × 32 = (–32)

First four terms of given series are 4, –8 , 16 ,–32