NCR Corporation Progression Questions with Solutions

Answer : D
Explanation
Here a = 14, d = (9 - 14) = -5.
∴ T₁₂ = a + (12 -1) d = a + 11d = 14 + 11 * (-5) = -41.

Answer : B
Explanation
Given series is an A.P. in which a=72 and d = (63-72)= -9
Let the nth term be 0. Then a+ (n-1) d =0 ⇒ 72+ (n-1)*(-9) = 0 ⇒ 72-9n +9 =0
⇒ 9n =81 ⇒ n = 9
∴ 9th term is the given series is 0.

Answer : B
Explanation
LCM of 4, 5, 6 = 2*2*5*3 = 60
So, each one must be divisible by 60.
Requisite numbers are 240, 300, 360, 420, 480 and 540.
Their no. is 6.

Answer : D
Explanation
This is an A.P. in which a = 2 and d = (5-2) =3.
∴ Next number= (8+3) =11.

Answer : B
Explanation
Since a, (a-2), 3a are in A.P. , we have
(a-2) - a = 3a - (a-2) ⇒ 2a +2 = -2 ⇒ 2a = -4 ⇒ a = -2.

Answer : B
Explanation
Let Tn = 1280. Then ar (ⁿ⁻ⁱ) = 1280 ⇒ Here a= 5 and r =2
∴ 5*2(ⁿ⁻ⁱ) =1280 ⇒256 = 2(ⁿ⁻ⁱ) =2⁸ ⇒ n-1 =8 ⇒ n= 9

Answer : C
Explanation
Let the number be a/r, a, ar.
Then, a/r* a*ar = 512 ⇒ a3 =83 ⇒ a =8
8/r+8+8r =28 ⇒ 8/r +8r =20 ⇒ 2/r+2r =5
⇒ 2r2-5r+2 =0 ⇒ 2r2-4r-r+2 =0
⇒ 2r (r-2)- (r-2)=0 ⇒ (r-2)(2r-1) =0
⇒r = 2 or r = 1/2
Numbers are 4,8, 16.

Answer : D
Explanation
Let the required value be Rs a.
Also d= 400, n = 8 and Sn = 48000.
Sn = n/2 [2a + (n -1) d] ⇒ 8/2 *[2a + 7 *400] = 48000
⇒ 2a + 2800 = 12000⇒ 2a=9200⇒ a = 4600

Answer : A
Explanation
(12+ 32+52+..........+192) = (12+22+32+42+52+........+182+192) - (22+42+62+.........+182)
= {19*(19+1) (38+1)/6} - (1*22+22*22+22*32+22*42+???. +22*92)
= 2470 -22*{12+22+32+... +92} = 2470 - (4*9*10*19)/6= (2470-1140) = 1330