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Mathematics

The sequence 2, 9, 16, ….. is given.

(a) Identify if the given sequence is an AP or a GP. Give reasons to support your answer.

(b) Find the 20th term of the sequence.

(c) Find the difference between the sum of its first 22 and 25 terms.

(d) Is 102 a term of this sequence?

(e) If ‘k’ is added to each of the above terms, will the new sequence be in A.P. or G.P.?

A.P.

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Answer

(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,

an = 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,

Sn = n2[2a+(n−1)d]\dfrac{n}{2}[2a + (n - 1)d]

Substituting values we get :

Sum upto 22 terms :

S22=222[2×2+(22−1)×7]=11×[4+21×7]=11×[4+147]=11×151=1661.S_{22} = \dfrac{22}{2}[2 \times 2 + (22 - 1) \times 7] \\[1em] = 11 \times [4 + 21 \times 7] \\[1em] = 11 \times [4 + 147] \\[1em] = 11 \times 151 \\[1em] = 1661.

Sum upto 25 terms :

S25=252[2×2+(25−1)×7]=252×[4+24×7]=252×[4+168]=252×172=25×86=2150.S_{25} = \dfrac{25}{2}[2 \times 2 + (25 - 1) \times 7] \\[1em] = \dfrac{25}{2} \times [4 + 24 \times 7] \\[1em] = \dfrac{25}{2} \times [4 + 168] \\[1em] = \dfrac{25}{2} \times 172 \\[1em] = 25 \times 86 \\[1em] = 2150.

S25 - S22 = 2150 - 1661 = 489.

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

(d) Let nth term be 102.

⇒ an = a + (n - 1)d

⇒ 102 = 2 + 7(n - 1)

⇒ 102 = 2 + 7n - 7

⇒ 102 = 7n - 5

⇒ 102 + 5 = 7n

⇒ 7n = 107

⇒ n = 1077=1527\dfrac{107}{7} = 15\dfrac{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.

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