Question

The sequence 2, 9, 16, ...... is given.

1. Identify if the given sequence is an AP or a GP. Give reasons to support your answer.
2. Find the 20th term of the sequence.
3. Find the difference between the sum of its first 22 and 25 terms.
4. Is the term 102 belong to this sequencе?
5. If ‘k’ is added to each of the above terms, will the new sequence be in A.P. or G.P.?

Answer 1

a. Difference between common terms:

16 – 9

= 9 – 2 

= 7

Since, difference between common terms are equal.

Hence, the given sequence is an A.P.

b. By formula,

a_n = a + (n – 1)d

a₂₀ = 2 + (20 – 1) × 7

= 2 + 19 × 7

= 2 + 133

= 135

Hence, 20th term of the sequence = 135.

c. By formula,

`S_n = n/2 [2a + (n - 1)d]`

Substituting values we get:

Sum upto 22 terms:

`S_22 = 22/2 [2 xx 2 + (22 - 1) xx 7]`

= 11 × [4 + 21 × 7]

= 11 × [4 + 147]

= 11 × 151

= 1661

Sum upto 25 terms:

`S_25 = 25/2 [2 xx 2 + (25 - 1) xx 7]`

= `25/2 xx [4 + 24 xx 7]`

= `25/2 xx [4 + 168]`

= `25/2 xx 172`

= 25 × 86

= 2150

S₂₅ – S₂₂ = 2150 – 1661

= 489

Hence, difference between the sum of its first 22 and 25 terms = 489.

d. Let n^th term be 102.

⇒ a_n = a + (n – 1)d

⇒ 102 = 2 + 7(n – 1)

⇒ 102 = 2 + 7n – 7

⇒ 102 = 7n – 5

⇒ 102 + 5 = 7n

⇒ 7n = 107

⇒ n = `107/7`

= `15 2/7`

Since, n cannot be in fraction.

Hence, 102 is not the term of the sequence.

e. If k is added to each term.

Sequence: 2 + k, 9 + k, 16 + k,.............

The common difference between terms is still equal to 7.

Hence, sequence is in A.P.