Practice question on Progression for IITJEE NDA and AIRFORCE 31-45

Progression

---

Question no. 31

If a,b,c are distinct such that ab+bc+ca≠0 and in AP, then a²(b+c) , b²(a+c) , c²(b+a) is

looks_one AP

looks_two GP

looks_3 HP

looks_4 none of these

Option looks_one AP

Solution :
a ,b ,c are in A.P.

Question no. 32

The sum of the series: 1² + 3² + 5² + .......... n terms is:

looks_one (4n²-1)

looks_two (4n-1)

looks_3 (4n+1)

looks_4(4n²+1)

Option looks_one (4n²-1)

Solution :

Question no. 33

If A and G are AM and GM between two number a and b, then roots of the equation: x² - 2Ax + G²=0 are:

looks_one a, 2b

looks_two 2a, b

looks_3 a, b

looks_4 2a, 2b

Option looks_3 a, b

Solution :
A = (a+b)/2 , G = √ab
x² -(a+b)x +ab = 0
a, b are roots of equation

Question no. 34

If the p^th ,q^th and r^th term of an AP are a,b and c respectively, then the value of [a(q-r)+b(r-p)+c(p-q)]=

looks_one 1

looks_two -1

looks_3 0

looks_4 1/2

Option looks_3 0

Solution :
A_p = a = A + (p-1)D
A_q = b = A + (q-1)D
A_r = c = A +(r-1)D
[a(q-r)+b(r-p)+c(p-q)] = (q-r)(A + (p-1)D) + (r-p)(A + (q-1)D) + (p-q)(A +(r-1)D) = 0

Question no. 35

The interior angles of a poygon are in AP. If the smallest angles be 120^o and the common difference be 5^o, then the number of sides is

looks_one 8

looks_two 10

looks_3 9

looks_4 6

Option looks_3 9

Solution :
No. of sides = n
sum of interior angle = (n-2)180^o
A = 120⁰ , d = 5^o
n{120^o*2 +(n-1)5^o }/2 = (n-2)180^o
5n² -125n + 720 = 0
n² -16n -9n + 144 = 0
n= 16 , 9

Question no. 36

If a₁, a₂, a₃, a₄,......... a_n are in AP, where a_i >0 for all i, then the value of

looks_one

looks_two

looks_3

looks_4

Option looks_one

Solution :
let d is the difference of AP

Question no. 37

If log_√3x + log_⁴√3x + log_⁶√3x ......... + log_¹⁶√3x =36 is

looks_one x=3

looks_two x=4√3

looks_3 x=9

looks_4 x=√3

Option looks_4 x=√3

Solution :
log_√3x + log_⁴√3x + log_⁶√3x ......... + log_¹⁶√3x =36
2log₃x + 4log₃x + 6log₃x ...... + 16log₃x = 36
log₃x[2 + 4 + 6 +... + 16] = 36
log₃x[72] = 36
log₃x = 1/2
x = √3

Question no. 38

A series whose n^th term is (n/x) + y , the sum of r terms will be

looks_one {r(r+1)/2x} +ry

looks_two {r(r-1)/2x}

looks_3 {r(r-1)/2x} -ry

looks_4 {r(r+1)/2y} -rx

Option looks_one {r(r+1)/2x} +ry

Solution :

Question no. 39

If ^x√a = ^y√b = ^z√c and a, b, c are in G.P., then x , y , z are in

looks_one AP

looks_two HP

looks_3 GP

looks_4 none of these

Option looks_one AP

Solution :
^x√a = ^y√b = ^z√c

Question no. 40

Find the sum of first n terms of the series : 3 + 7 + 13 + 21 + 31 ...............

looks_one (n/2)(n² +3n+5)

looks_two (n/4)(n² +3n+5)

looks_3 (n/3)(n² +3n+5)

looks_4 (n/6)(n² +3n+5)

Option looks_3 (n/3)(n² +3n+5)

Solution :
S = 3 + 7 + 13 + 21 + 31 + ........ n terms (1)
S =     3 + 7 + 13 + 21 + .... n terms (2)
subtract (2) from (1)
0 = 3 + 4 + 6 + 8 + 10 +....... - a_n
a_n = 3 + (4 + 6 + 8 + 10 +........(n-1)terms)
a_n = 3 + (n-1)(n+2) = n² + n + 1
S_n = Σn² + Σn + Σ1

Question no. 41

Find the sum of first infinity terms of the series :

looks_one 35/4

looks_two 5/16

looks_3 7/16

looks_4 35/16

option

very soon 11

Question no. 42

Find the sum of first n terms of the series :

looks_one (2n² +9n+13)

looks_two (2n² +9n+13)

looks_3 (2n² +6n+13)

looks_4 (2n² +6n+13)

option

very soon 12

Question no. 43

If a₁, a₂, a₃, a₄,......... a_n are in HP, then a₁a₂ + a₂a₃ +..........+ a_n-1a_n=ka₁a_n , where k is

looks_one n-1

looks_two n

looks_3 1

looks_4 1/n

option

very son 13

Question no. 44

If , then which of the following true?

looks_one a+c = 2b

looks_two

looks_3

looks_4 a-c = 2b

Option looks_two

Solution :

Question no. 45

If the sum of three numbers in HP is 26 and the sum of their reciprocal is 3/8 , then the numbers are :

looks_one 6,8,10

looks_two 6,9,12

looks_3 6,8,12

looks_4 8,10,12

option looks_3 6,8,12

Let the three terms of H.P. are

a = 1/8 ⇒ second term is 8.
Put the value of a in first equation

Previous Next