Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1
Solution & Step-by-Step Answer:

(ii) 1, -5, 25, -125, ………
Solution:

(iii)
Solution:

(iv) 3, 4, 5, 6, ……
Solution:

(v) 7, 14, 21, 28, ……
Solution:

Solution & Step-by-Step Answer:

(ii) If a = , r = 3, find t6.
Solution:

(iii) If r = -3 and t6= 1701, find a.
Solution:

(iv) If a = , t6= 162, find r.
Solution:

Solution & Step-by-Step Answer:

Solution & Step-by-Step Answer:

Solution & Step-by-Step Answer:


Solution & Step-by-Step Answer:
Let the three numbers in G. P. be , a, ar. According to the given conditions, When a = 6, r = 2, = 3, a = 6, ar = 12 Hence, the three numbers in G.P. are 12, 6, 3 or 3, 6, 12.


Check:
If sum of the three numbers is 21 and sum of their squares is 189, then our answer is correct.
Sum of the numbers = 12 + 6 + 3 = 21
Sum of the squares of the numbers = 122+ 62+ 32
= 144 + 36 + 9
= 189
Thus, our answer is correct.
Solution & Step-by-Step Answer:
Let the four numbers in G.P. be According to the given conditions,


Solution & Step-by-Step Answer:
Let the five numbers in G. P. be According to the given conditions, Hence, the five numbers in G.P. are 1, 2, 4, 8, 16 or 1, -2, 4, -8, 16.

Solution & Step-by-Step Answer:

Solution & Step-by-Step Answer:
p, q, r, s are in G.P. ∴ Let = k ∴ q = pk, r = qk, s = rk We have to prove that p + q, q + r, r + s are in G.P. i.e., to prove that ∴ p + q, q + r, r + s are in G.P.

Solution & Step-by-Step Answer:
Since the number of bacteria in culture doubles every hour, increase in number of bacteria after every hour is in G.P. ∴ a = 50, r = = 2 tn = arn-1 To find the number of bacteria at the end of the 5th hour. (i.e., to find the number of bacteria at the beginning of the 6th hour, i.e., to find t6.) ∴ t6 = ar5 = 50 × (25) = 50 × 32 = 1600
Solution & Step-by-Step Answer:
Since the ball rebounds of the height it has fallen, the height in successive bounce is in G.P. 1st height in the bounce = 80 ×

Solution & Step-by-Step Answer:
(i) 3, x and x + 6 are in G. P. x2 = 3x + 18 x2 – 3x – 18 = 0 (x – 6) (x + 3) = 0 x = 6, -3

Solution & Step-by-Step Answer:
a = 200, r = 1 + = Mosquitoes at the end of 1st year = 200 × (i) Number of mosquitoes after 3 years = 200 × = 200 = 200 (1.1)3
(ii) Number of mosquitoes after 10 years = 200 (1.1)10
(iii) Number of mosquitoes after n years = 200 (1.1)n
Solution & Step-by-Step Answer:
(i) x – 6, 2x and x are in Geometric progression. ∴ 4x2 = x2(x – 6) 4 = x – 6 x = 10
(ii) t1= x – 6 = 10 – 6 = 4
