Sequences and Series

More Topic from Mathematics

Question 1

Let space straight S subscript 1 space equals space sum from straight j space equals 1 to 10 of space straight j space left parenthesis negative 1 right parenthesis space to the power of 10 straight C subscript straight j space comma space straight S subscript 2 space equals space sum from straight j space equals 1 to 10 of space straight j to the power of 10 straight C subscript straight j space and space straight S subscript 3 space equals sum from straight j space equals 1 to 10 of space straight j squared space to the power of 10 straight C subscript straight j

Statement-1: S3 = 55 × 29.
Statement-2: S1 = 90 × 28 and S2 = 10 × 28.

  • Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

  • Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1. 

  • Statement-1 is true, Statement-2 is false.

  • Statement-1 is false, Statement-2 is true.

Solution

C.

Statement-1 is true, Statement-2 is false.

straight S subscript 1 space equals space straight capital sigma with straight i space equals space 1 below and 10 on top space straight j space left parenthesis straight j minus 1 right parenthesis fraction numerator 10 factorial over denominator straight j space left parenthesis straight j minus 1 right parenthesis left parenthesis straight j minus 2 right parenthesis factorial left parenthesis 10 minus straight j right parenthesis factorial end fraction
space equals space 90 sum from straight j space equals 2 to 10 of space fraction numerator 8 factorial over denominator left parenthesis straight j minus 2 right parenthesis factorial left parenthesis 8 minus left parenthesis straight j minus 2 right parenthesis right parenthesis factorial end fraction space equals space 90.2 to the power of 8
space and space straight S subscript 2 space equals space sum from straight i space equals 1 to 10 of fraction numerator 9 factorial over denominator straight j left parenthesis left parenthesis straight j minus 1 right parenthesis factorial space left parenthesis 9 minus left parenthesis straight j minus 1 right parenthesis right parenthesis factorial end fraction
space equals space 10 sum from straight j space equals 1 to 10 of space fraction numerator 9 factorial over denominator left parenthesis straight j minus 1 right parenthesis factorial left parenthesis 9 minus left parenthesis straight j minus 1 right parenthesis right parenthesis factorial end fraction space equals space 10.2 to the power of 9
Also comma space straight S subscript 3 space equals space 10 sum from straight j space equals 1 to 10 of space left square bracket space straight j space left parenthesis straight j minus 1 right parenthesis space plus straight j right square bracket fraction numerator 10 factorial over denominator straight j factorial left parenthesis 10 minus straight j right parenthesis factorial end fraction
space equals space sum from straight j space equals 1 to 10 of space space straight j space left parenthesis straight j minus 1 right parenthesis to the power of 10 straight C subscript straight i space equals space sum from straight j space equals 1 to 10 of space straight j to the power of 10 straight C subscript straight j
space equals space 90.2 to the power of 8 space plus space 10.2 to the power of 8
space equals space 90.2 to the power of 8 space plus space 20.2 to the power of 8
space equals space 1110.2 to the power of 8 space equals space 56.2 to the power of 8
Question 2

A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after

  • 18 Months

  • 19 Months

  • 20 Months

  • 21 Months

Solution

D.

21 Months

a = Rs. 200
d = Rs. 40
savings in first two months = Rs. 400
remained savings = 200 + 240 + 280 + ..... upto n terms =
n/2[400 + (n -1)40] = 11040 - 400
200n + 20n2 - 20n = 10640
20n2 + 180 n - 10640 = 0
n2 + 9n - 532 = 0
(n + 28) (n - 19) = 0
n = 19
∴ no. of months = 19 + 2 = 21

Question 3

A person is to count 4500 currency notes. Let a denote the number of notes he counts in the nth minute. If a1 = a2 = ... = a10 = 150 and a10, a11, ...are in an AP with common difference -2, then the time taken by him to count all notes is

  • 24 min

  • 34 min

  • 125 min

  • 135 min

Solution

B.

34 min

Let the first term of an AP be a and common difference be d and number of terms be n, then 
tn = a + (n-1)d and Sn = n/2 [ 2a + (n-1)d]
Number of notes that the person counts in 10 min = 10 x 150 = 1500
Since a10, a11, a12, .... are in AP in with common difference -2
Let n be the time has taken to count remaining 3000 notes than
n/2[2 x 148 + (n-1) x -2] = 3000
⇒ n2-149n +3000 = 0 
⇒ (n-24)n-125) = 0 
n = 24, 125
Then, the total time taken by the person to count all notes = 10 +24 = 34 min
n = 125 is discarded as putting n = 125
an = 148 + (125-1)(-2)
= 148 - 124 x 2 = 148-248 = -100
⇒ Number of notes cannot be negative.

Question 4

For any three positive real numbers a, b and c, (25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c), Then

  • b , c and a are in G.P

  • b, c and a are in A.P

  • a, b and c are in A.P

  • a, b and c are in G.P

Solution

B.

b, c and a are in A.P

225a2 + 9b2 + 25c2 – 75ac – 45ab – 15bc = 0
(15a)2 + (3b)2 + (5c)2 – (15a)(3b) – (3b)(5c) – (15a) (5c) = 0
1/2[(15a – 3b)2 + (3b – 5c)2 + (5c – 15a)2] = 0
15a = 3b , 3b = 5c , 5c = 15a
5a = b , 3b = 5c , c = 3a
a/1 = b/5 = c/3
a = λ, b = 5λ, c = 3λ
a, c, b are in AP

Question 5

If straight x space equals space sum from straight n equals space 0 to infinity of straight a to the power of straight n comma space straight y space equals space sum from straight n equals 0 to infinity of straight b to the power of straight n comma space sum from straight n equals 0 to infinity of straight c to the power of straight n where a, b, c are in A.P. and |a| < 1, |b|<1, |c|< 1, then x, y, z are in

  • G.P.

  • A.P.

  • Arithmetic − Geometric Progression

  • H.P.

Solution

D.

H.P.

straight x space equals sum from straight n equals 0 to infinity of space straight a to the power of straight n space equals space fraction numerator 1 over denominator 1 minus straight a end fraction space space space space space space space space space straight a equals space 1 minus 1 over straight x
straight y space equals sum from straight n equals 0 to infinity of space straight b to the power of straight n space space space equals fraction numerator 1 over denominator 1 minus straight b end fraction space space space space space space space space straight b equals space space 1 minus 1 over straight y
straight z space equals space sum from straight n equals space 0 space to infinity of straight c to the power of straight n space equals space fraction numerator 1 over denominator 1 minus straight c end fraction space space space space space space space space space straight c space equals space 1 minus 1 over straight z
straight a comma straight b comma space straight c space are space in space straight A. straight P.
2 straight b space equals space straight a plus straight c
2 open parentheses 1 minus 1 over straight y close parentheses space equals 1 minus 1 over straight x plus 1 minus 1 over straight y
2 over straight y space equals space 1 over straight x plus 1 over straight z
rightwards double arrow space straight x comma straight y comma straight z space are space in space straight H. straight P.