Q: [CBSE Board Exam 2024 - 3 Marks] Find the sum of the series 5 + 55 + 555 + … to n terms.

Each term: 5(1 + 11 + 111 + … ) = 5/9 (9 + 99 + 999 + …)= 5/9 [(10−1) + (100−1) + (1000−1) + …]= 5/9 [(10 + 10² + … + 10ⁿ) − n]= 5/9 [10(10ⁿ − 1)/9 − n]S(n) = 50/81 (10ⁿ − 1) − 5n/9.

Q: [CBSE Board Exam 2024 - 2 Marks] The sum of the first three terms of a GP is 13/12 and their product is −1. Find the terms.

Let terms be a/r, a, ar. Product = a³ = −1 ⇒ a = −1.Sum = −1/r − 1 − r = 13/12 ⇒ −1/r − r = 25/12 ⇒ 12r² + 25r + 12 = 0.r = (−25 ± 7)/24 ⇒ r = −3/4 or r = −4/3.For r = −3/4: terms are 4/3, −1, 3/4. For r = −4/3: terms are 3/4, −1, 4/3.Terms: 4/3, −1, 3/4 (in either order).

Question 1

[CBSE Board Exam 2024 - 3 Marks] Find the sum of the series 5 + 55 + 555 + … to n terms.

Accepted Answer

Each term: 5(1 + 11 + 111 + … ) = 5/9 (9 + 99 + 999 + …)
= 5/9 [(10−1) + (100−1) + (1000−1) + …]
= 5/9 [(10 + 10² + … + 10ⁿ) − n]
= 5/9 [10(10ⁿ − 1)/9 − n]
S(n) = 50/81 (10ⁿ − 1) − 5n/9.

Question 2

[CBSE Board Exam 2024 - 2 Marks] The sum of the first three terms of a GP is 13/12 and their product is −1. Find the terms.

Accepted Answer

Let terms be a/r, a, ar. Product = a³ = −1 ⇒ a = −1.
Sum = −1/r − 1 − r = 13/12 ⇒ −1/r − r = 25/12 ⇒ 12r² + 25r + 12 = 0.
r = (−25 ± 7)/24 ⇒ r = −3/4 or r = −4/3.
For r = −3/4: terms are 4/3, −1, 3/4. For r = −4/3: terms are 3/4, −1, 4/3.
Terms: 4/3, −1, 3/4 (in either order).

Question 3

[CBSE Board Exam 2023 - 3 Marks] If a, b, c are in AP, prove that 1/bc, 1/ca, 1/ab are also in AP.

Accepted Answer

a, b, c in AP ⇒ 2b = a + c.To prove 1/bc, 1/ca, 1/ab in AP: show 2/ca = 1/bc + 1/ab.RHS = (a + c)/abc = 2b/abc = 2/ac = LHS. Proved.

Sequences and Series - Previous Year Questions