Arithmetic Progression — Analytical & Application Based Questions

(a) Write the nth term (T_n) of an Arithmetic Progression (A.P.) consisting of all whole numbers which are divisible by 3 and 7.

(b) How many of these are two-digit numbers? Write them.

(c) Find the sum of first 10 terms of this A.P.

Answer

(a) A.P. = 21, 42, 63, .........

The above sequence is an A.P. with first term (a) = 21 and common difference (d) = 21.

By formula,

T_n = a + (n - 1)d

= 21 + (n - 1)21

= 21 + 21n - 21

= 21n.

Hence, the n^th term = 21n.

(b) A.P. = 21, 42, 63, 84, 105, ........

Hence, there are four two digit numbers i.e. 21, 42, 63, 84 in the A.P.

(c) By formula,

Sum of A.P. = $\dfrac{n}{2}(a + a_n)$

$\text{Sum of first 10 terms of A.P.} = \dfrac{10}{2}(a + a_{10}) \\[1em] = 5[a + a + (n - 1)d] \\[1em] = 5[2a + (n - 1)d] \\[1em] = 5[2 \times 21 + (10 - 1) \times 21] \\[1em] = 5[42 + 9 \times 21] \\[1em] = 5[42 + 189] \\[1em] = 5 \times 231 \\[1em] = 1155.$

Hence, sum of first 10 terms of the A.P. = 1155.

The sum of a certain number of terms of the Arithmetic Progression (A.P.) 20, 17, 14, ....... is 65. Find the:

(a) number of terms.

(b) last term.

Answer

(a) Let no. of terms be n.

By formula,

Sum of A.P. = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow 65 = \dfrac{n}{2}[2 \times 20 + (n - 1) \times (-3)] \\[1em] \Rightarrow 65 = \dfrac{n}{2}[40 - 3n + 3] \\[1em] \Rightarrow 65 = \dfrac{n}{2}[43 - 3n] \\[1em] \Rightarrow 65 \times 2 = n(43 - 3n) \\[1em] \Rightarrow 130 = 43n - 3n^2 \\[1em] \Rightarrow 3n^2 - 43n + 130 = 0 \\[1em] \Rightarrow 3n^2 - 30n - 13n + 130 = 0 \\[1em] \Rightarrow 3n(n - 10) - 13(n - 10) = 0 \\[1em] \Rightarrow (3n - 13)(n - 10) = 0 \\[1em] \Rightarrow 3n - 13 = 0 \text{ or } n - 10 = 0 \\[1em] \Rightarrow 3n = 13 \text{ or } n = 10 \\[1em] \Rightarrow n = \dfrac{13}{3} \text{ or } n = 10.$

Since, no. of terms cannot be in fraction.

∴ n = 10.

Hence, no. of terms = 10.

(b) By formula,

Last term (l) = a + (n - 1)d

= 20 + (10 - 1) × (-3)

= 20 + 9 × (-3)

= 20 - 27

= -7.

Hence, last term = -7.

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.?

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,

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

Substituting values we get :

Sum upto 22 terms :

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

Sum upto 25 terms :

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

S₂₅ - S₂₂ = 2150 - 1661 = 489.

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

(d) Let 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 = $\dfrac{107}{7} = 15\dfrac{2}{7}$.

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.