Practice Question On Progression for IITJEE , NDA And AIRFORCE 16-30(SET 2 )

Progression

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Question no. 16

The sum of 24 terms of the series √2 + √8 + √18 + √32 .... is

looks_one 300

looks_two 300√2

looks_3 200√2

looks_4 none of these

Option looks_two 300√2 is correct answer

Given series is ,√2 + √8 + √18 + √32
⇒ it can be write down like √2 + 2√2 + 3√2 + 4√2 ,....
⇒ it is clearly observed that given series is in A.P. with
⇒ a= √2 d =√2
⇒ sum of first 24 terms = (n/2) { 2a+ (n-1)d }
= (24/2) { 2√2 + (24-1)√2 }
= 12 { 2√2 +23√2 }
= 12 { 25√2 }
= 300√2

Question no. 17

If log₃2, log₃(2^x-5) and log₃(2^x-7/2) are in A.P. , then the value of x is

looks_one 1, 1/2

looks_two 1, 1/3

looks_3 1, 3/2

looks_4 none of these

option looks_4 none of these

Given log₃2, log₃(2^x-5) and log₃(2^x-7/2) are in A.P.
it means , 2log₃(2^x-5) = log₃2 + log₃(2^x-7/2)
⇒log₃(2^x-5)² = log₃( 2.(2^x-7/2)) [log a + log b = log ab ]
⇒(2^x-5)² = 2.(2^x-7/2)
⇒ let 2^x= Y
⇒ (Y-5)² = 2.( Y -7/2)
⇒ Y² + 25 - 10Y = 2Y - 7
⇒ Y² - 12Y +32 = 0
⇒ Y² - 8Y -4Y + 32=0
⇒ (Y- 8)(Y-4) = 0
⇒ Y = 8 , 4
(i) 2^x = 8
x=3
(ii) 2^x = 4
x = 2

Question no. 18

What is the sum of the series 0.5 + 0.55 + 0.555....... to n terms?

looks_one

looks_two

looks_3

looks_4

Option looks_4

Solution :

Question no. 19

If the sum of an infinite G.P. be 9 and the sum of first two terms be 5, then, the common ratio is

looks_one 1/3

looks_two 3/2

looks_3 3/4

looks_4 2/3

Optionlooks_4 2/3

Solution :
a/(1-r) = 9 ⇒ a= 9(1-r)
a+ar = 5
a(1+r) = 5
9 (1- r² ) = 5
1-r² = 5/9
r² = 4/9 ⇒ r= 2/3

Question no. 20

The sum of 100 terms of the series 0.9 + 0.09 + 0.009 ......... will be

looks_one 1-

looks_two 1+

looks_3 1-

looks_4 none of these

Option looks_one 1-

Solution :

Question no. 21

If the roots of equation ax³ + bx² + cx+d=0 are in GP , then

looks_one c³a=b³d

looks_two ca³=bd³

looks_3 a³b=c³d

looks_4 ab³=cd³

Option looks_one c³a=b³d

Solution :

Question no. 22

If log_xa , a^x/2 ,and log_bx are in G.P. then x=

looks_one -log(log_ba)

looks_two log_a(log_ab)

looks_3 log_a(log_ea)- log_a(log_eb)

looks_4 log_a(log_eb)- log_a(log_ea)

Option looks_3 log_a(log_ea)- log_a(log_eb)

Solution :
log_xa , a^x/2 ,and log_bx are in G.P
(a^x/2)² = log_xa . log_bx
a^x = log_ea.log_ex/(log_ex.log_eb)
a^x = log_ea/log_eb
apply log_a
log_aa^x = log_a[log_ea - log_eb]
a = log_a.log_ea - log_a.log_eb

Question no. 23

If a,b,c are the p^th ,q^th and r^th terms of GP, then (b/a)^r(c/b)^p(a/c)^q is equal to

looks_one 1

looks_two a^pb^qc^r

looks_3 a^qb^rc^p

looks_4 a^rb^pc^q

Option looks_one 1

Solution :

Question no. 24

If x, y, z are in A.P., then y+z, z+x, x+y are in

looks_one AP

looks_two HP

looks_3 GP

looks_4 none of these

Option looks_one AP

Solution :
x, y ,z are in AP
2y =x+z
If terms(already in A.P.) are in addition , it must be in AP.
Let's check
y+z , z+x ,x+y
To prove : 2(z+x) = y+z + x+y
y+z + x+y = 2y + x + z = x+z + x+z
= 2x+ 2z
Following terms are in AP

Question no. 25

Find the sum of first n terms of the series : 1(1)! + 2(2)! + 3(3)! + ...............

looks_one n! -1

looks_two (n+1)! +1

looks_3 (n+1)! -1

looks_4 n! + 1

option looks_3 (n+1)! -1

Solution :
1(1)! + 2(2)! + 3(3)! + ...............
a_n = n.(n)! = (n+1-1)n! = (n+1)n! -n! = (n+1)! - n!
a_n = (n+1)! - n!
S_n = ∑ (n+1)! - n!
= (2! - 1 !) + (3! - 2!) + ..........(n! - (n-1)!) + ((n+1)! - n!) = (n+1)! - 1

Question no. 26

Find the sum of first n terms of the series :

looks_one 1/(3n-1)

looks_two 1/(3n+1)

looks_3 n/(3n-1)

looks_4 n/(3n+1)

option looks_4 n/(3n+1)

Solution :

Question no. 27

In an AP, the ratio of sum of m terms and n terms is m² : n². What is the ratio of mth and nth term ?

looks_one (m-1)/(n-1)

looks_two (n-1)/(m-1)

looks_3 (2m-1)/(2n-1)

looks_4 (2n-1)/(2m-1)

option looks_3 (2m-1)/(2n-1)

Solution :

Question no. 28

find the S = 1 + 4 + 10 + 22 + 46 +.........to n terms

looks_one 3.2ⁿ +2n -3

looks_two 3.2ⁿ +2n + 3

looks_3 3.2ⁿ -2n -3

looks_4 3.2ⁿ -2n +3

option looks_3 3.2ⁿ -2n -3

Solution :
S = 1 + 4 + 10 + 22 + 46 +.........a_n
S =      1 + 4 + 10 + 22 + 46 +.........a_n
0 = 1 + { 3 + 6 + 12 +....a_n -a_n-1} -a_n
a_n = 1 + { 3 + 6 + 12 +....a_n -a_n-1 }
a_n = 1 + 3 [2^n-1 -1] = 3.2^n-1 -2
S = ∑ 3.2^n-1 -2 = 1.5∑ 2ⁿ - 2n = 1.5 [2(2ⁿ - 1)] - 2n = 3.2ⁿ -3 - 2n

Question no. 29

If , then a, b,c are in

looks_one AP

looks_two GP

looks_3 HP

looks_4 none of these

Option looks_3 HP

Solution :

Question no. 30

The sums of n terms of A.P. are in the ratio 2n+3 : 6n+5, then the ratio of their 13th terms is

looks_one 53:155

looks_two 27:77

looks_3 29:83

looks_4 31:89

Option looks_one 53:155

Solution :

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