Practice question on progression (A.P. , G.P. and H.P.) for iit-jee , NDA and Airforce

Progression

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Question no. 1

What is the 15^th term of the series 3, 7 , 13 , 21 , 31 ....?

looks_one 205

looks_two 225

looks_3 238

looks_4 241

looks_4 241

Solution :
Given series is 3, 7 ,13 ,21 , 31 .....
as it is observed that given series is not in A.P. , G.P. and H.P.
But difference of terms 7-3=4 , 13-7=6 , 21-13= 8 31-21 =10 are in A.P.
if differences of term are in A.P. , there is formula of nth term in this type of series.
t_n = n²a + bn + c
t₁= a(1)² + b(1) + c =a + b + c = 3
t₂= a(2)² + b(2) + c = 4a + 2b + c = 7
t₃= a(3)² + b(3) + c = 9a + 3b + c = 13
solve all three equation ,find the value of a,b,c
a= 1
b= 1
c= 1
now, tn = n² + n + 1
t₁₅ = (15)² + (15) + 1 = 225 +16
= 241

Question no. 2

If the n^th term of an A.P. is 2n-1, what is the sum of upto n terms ?

looks_one n²

looks_two n²-1

looks_3 n²+1

looks_4 0.5n(n+1)

Option looks_one n²

Solution :
Given a_n = 2n - 1
a₁ = 2(1) - 1 = 1
now , sum of nth term = (n/2) ( a + a_n)
= (n/2) ( 1 + 2n - 1 )
= (n/2)(2n) = n²

Question no. 3

If the three observation are 3,-6 and -6. Then what is their harmonic mean?

looks_one 0

looks_two -3

looks_3 -0.5

looks_4 ∞

Option looks_4 ∞

Solution :
let H.M. is x
there are 3 observation , so
3/x = 1/a + 1/b + 1/c
3/x = 1/3 + 1/(-6) + 1/(-6)
3/x = 1/3 - 1/3
3/x = 0
x = 3/0 = infinity

Question no. 4

Sum of first n natural numbers is given by n(n+1)/2. what is the geometric means of the series 1 , 2, 4, 8.....2ⁿ ?

looks_one 2ⁿ

looks_two 2^n/2

looks_3 2^1/2

looks_4 2^n-1

Option looks_two 2^n/2

Solution :
G.M. = ⁿ√(a₁a₂....a_n)
In the given series , there are n+1 terms
G.M. = ⁿ⁺¹√( 1.2.4 .8....2ⁿ)
= ⁿ⁺¹√ (2 ^n(n+1)/2 )
= 2 ^n/2

Question no. 5

If the number of terms of an A.P. is (2n+1), then what is the ratio of the sum of odd terms to the even terms?

looks_one n/(n+1)

looks_two n²/(n+1)

looks_3 (n+1)/n

looks_4 (n+1)/2n

Option looks_3 (n+1)/n

Solution :
Given : no. of terms of an A.P. is 2n+1
let the A.P. is a , a+d , a+2d ....... a+(2n)d where a is first term and d is difference
even terms - a+d , a+3d ........a +(2n-1)d , and total number of even terms will be n.
odd terms - a, a+2d ...... a+(2n)d , the total number of odd terms will be n+1.
S_even= (n/2)(a+d + a+ (2n-1)d) = (n/2)(2a + 2nd) -(1)
S_even= ((n+1)/2)(a + a + (2n)d) = ((n+1)/2)(2a + 2nd) -(2)
divide (2) by (1)
S_odd / S_even= n+1 / n

Question no. 6

If the sum of terms of an A.P. is n² -2n , then what is the n^th term?

looks_one 3n-n²

looks_two 2n-3

looks_3 2n+3

looks_4 2n-5

Option looks_two 2n-3

Solution :
There are two method od solving this question
Method one:
using direct formula
a_n = S_n - S_n-1
a_n= n² - 2n -{ (n-1)² - 2(n-1) }
a_n= n² - 2n - { n² + 1 - 2n -2n +2 }
a_n= n² - 2n - n² + 4n -3
= 2n -3
Second Method :
S₁ =sum of first one term = a(first term) = (1)² - 2(1) = -1
a= -1
S₂ =sum of first two term = a(first term) + (a+d)(second term) = (2)² - 2(2)
2a+ d= 0
2(-1) + d= 0
d= 2
a_n= a+(n-1)d
= -1 + (n-1)2 = 2n -3

Question no. 7

If the AM and GM of two numbers are 5 and 4 respectively, then what is the HM of those numbers?

looks_one 5/4

looks_two 16/5

looks_3 9/2

looks_4 9

Option looks_two 16/5

Solution :
Given :
A.M.= 5
G.M. = 4
H.M. = ?
H.M. = G.M.² / A.M.
= (4)² / 5
16/5

Question no. 8

The HM of two numbers is 21.6. If one of the number is 27, then what is the other number?

looks_one 162

looks_two 17.3

looks_3 18

looks_4 20

Option looks_3 18

Solution :
H.M. = 2ab/(a+b)
21.6 = 2 * 27 * b /(27 + b)
21.6 *(27+b) = 54b
583.2 + 21.6b = 54b
54b-21.6b= 583.2
32.4b= 583.2
b=18

Question no. 9

If x² , y², z² 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_two HP

Solution :
x² ,y² ,z² are in A.P.
it means 2y²= x² + z²
let's try for H.P.
if the y+z, z+x , x+y is in H.P.
2/(x+z) = 1/(y+z) + 1/(y+x)
= ( y+x+ y+z)/ (y+z)(y+x)
= (2y + x+z)/(y² + y(x+z) + zx)
= 2(2y + x+z)/(x² + z² + 2y(x+z) + 2zx)
= 2(2y + x+z)/((x+z)² + 2y(x+z))
= 2(2y + x+z)/((x+z)(2y +x +z))
= 2/(x+z)
it is in H.P.

Question no. 10

If x,2x+2,3x+3 are the first three terms of a GP, then what is the 4^TH term?

looks_one -27/2

looks_two 27/2

looks_3 33/2

looks_4 -33/2

Option looks_one -27/2

Solution :
Given x, 2x + 2 and 3x + 3 are the first three terms of G.P.
it means a= x (1)
ar= 2x+ 2 (2)
ar²= 3x + 3 (3)
ar.r = 3x+ 3
from (2) and (3)
ar²/ar = 3x + 3 / 2x+ 2
r= 3x+3 / 2x+ 2 (4)
from (1) and (2)
ar/a= 2x +2 / x
r= = 2x+ 2 /x (5)
equate (4) and (5)
3x+3/ 2x+ 2 = 2x+ 2 / x
3x² + 3x = 4x² + 4 + 8x
0= x² +5x + 4
0= x² +4x +x + 4
0= (x+4) (x+1)
x=-4 , -1
-1 can't be taken because it will make second and third term zero.
a=x= -4
ar=-6
r= 3/2
fourth term ar³= -4 * (27 /8 ) = -27/4

Question no. 11

If the AM and HM of two number are 27 and 12 respectively , then what is their GM equal to?

looks_one 12

looks_two 18

looks_3 24

looks_4 27

Option looks_two 18

Solution :
H.M. = G.M.² / A.M.
12 = G.M.² / 27
G.M.² = 27* 12 =324
G.M. = 18

Question no. 12

In an AP, the m^th term is 1/n and n^th term is 1/m. What is the (mn)^th term ?

looks_one 1/(mn)

looks_two m/n

looks_3 n/m

looks_4 1

Option looks_4 1

Solution :
a_m = 1/n
a_n = 1/m
a_m =a+ (m-1)d = 1/n (1)
a_n =a+ (n-1)d = 1/m (2)
solve (1) and (2) , find the value of a and d in terms of m and n.
a= 1/mn
d= 1/mn
Now , a_mn = a+ (mn-1)d
= 1/mn + (mn -1)/mn
= 1/mn + 1 -1/mn
= 1

Question no. 13

The 59^th term of an AP is 449 and the 449^th is 59. Which term is Equal to 0 (zero) ?

looks_one 501^st term

looks_two 508^th term

looks_3 502^nd term

looks_4 509^th term

Option looks_two 508^th term

Solution :
a⁵⁹ = a+ 58d =449 (1)
a⁴⁴⁹ = a+ 448d =59 (2)
solve (1) and (2) , find the value of a and d
a = 507 , d= -1
for a_n=0
a + (n-1)d =0
507 + (n-1 )(-1) = 0
507 -n + 1 =0
n=508

Question no. 14

If p times the pth term of an AP is q times the qth term, then what is the (p+q)th term equal to ?

looks_one p+q

looks_two pq

looks_3 1

looks_4 0

Option looks_4 0

Solution :
pa_p=qa_p
p ( a + (p-1)d ) = q ( a + (q-1)d )
pa + p²d -pd = qa + q²d -qd
pa - qa +p²d - q²d -pd +qd =0
a(p-q) +d(p² - q²) -d(p-q) =0
(p-q) { a + (p+q)d - d} = 0
a+ (p+q -1)d=0 (p -q can't be to equal to zero )
a_p+q = 0

Question no. 15

The geometric mean of three numbers was found as 6. It was subsequently found that , in this computation, a number 8 was wrongly read as 12. What is the correct GM ?

looks_one 4

looks_two ∛5

looks_3 2∛18

looks_4 none of these

Option looks_one 4

Solution :
let the three number are a/r, a ,ar
G.M. of three numbers = (a/r . a. ar )^1/3
6 = (a³)^1/3
6= a
Now , according to question , one of number is 12 ( which provide the G.M. 6 )
ar= 12
r= 2
but ar = 8
2a = 8
a= 4
G.M = (a/r. a. ar)^1/3 = a = 4

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