AP GP Questions for SSC CGL
Download SSC CGL AP & GP Questions PDF based on previous papers very useful for SSC CGL exams. AP & GP Questions for SSC exams.
Download AP GP Questions for SSC CGL
Get 200 SSC mocks for just Rs. 249. Enroll here
Take a free mock test for SSC CGL
Download SSC CGL Previous Papers PDF
More SSC CGL Important Questions and Answers PDF
Question 1: The 3rd and 9th term of an arithmetic progression are -8 and 10 respectively. What is the 16th term?
a) 34
b) 28
c) 25
d) 31
Question 2: What is the sum of the first 12 terms of an arithmetic progression if the first term is -19 and last term is 36?
a) 192
b) 230
c) 102
d) 214
Question 3: What is the sum of the first 12 terms of an arithmetic progression if the 3rd term is -13 and the 6th term is -4?
a) 67
b) 45
c) -30
d) -48
Question 4: What is the sum of the first 17 terms of an arithmetic progression if the first term is -20 and last term is 28?
a) 68
b) 156
c) 142
d) 242
Question 5: If the 3rd and the 5th term of an arithmetic progression are 13 and 21, what is the 13th term?
a) 53
b) 49
c) 57
d) 61
SSC CGL Previous Papers Download PDF
Question 6: The greatest number,that divides 43, 91 and 183 so as to leave the same remainder in each case, is
a) 9
b) 8
c) 4
d) 3
Question 7: The greatest number among $\sqrt{5}$,$\sqrt[3]{4}$,$\sqrt[5]{2}$,$\sqrt[7]{3}$ is
a) $\sqrt[3]{4}$
b) $\sqrt[7]{3}$
c) $\sqrt{5}$
d) $\sqrt[5]{2}$
Question 8: What is the greatest number which when divides 460, 491 and 553, leave 26 as remainderin each case?
a) 33
b) 27
c) 35
d) 31
Question 9: If the 10-digit number 897359y7x2 is divisible by 72, then what is the value of(3x-y), for the possible greatest value of y ?
a) 3
b) 8
c) 7
d) 5
Question 10: If the Arithmetic mean of 7, 5, 13, x and 9 is 10, then the value of x is
a) 10
b) 12
c) 14
d) 16
18000+ Questions – Free SSC Study Material
Answers & Solutions:
1) Answer (D)
Let the first term of an AP = $a$ and the common difference = $d$
3th term of AP = $A_3=a+2d=-8$ ———-(i)
9th term = $A_9=a+8d=10$ ——–(ii)
Subtracting equation (i) from (ii), we get :
=> $8d-2d=10-(-8)$
=> $6d=18$
=> $d=\frac{18}{6}=3$
Substituting it in equation (ii), => $a=10-8(3)=10-24=-14$
$\therefore$ 16th term = $A_{16}=a+15d$
= $-14+15(3)=-14+45=31$
=> Ans – (D)
2) Answer (C)
First term of AP = $a=-19$ and last term = $l=36$
Number of terms = $n=12$
Sum of AP = $\frac{n}{2}(a+l)$
= $\frac{12}{2}(-19+36)$
= $17 \times 6=102$
=> Ans – (C)
3) Answer (C)
$T_{3}$ = a + 2d = -13——-(1)
$T_{6}$ = a + 5d = -4——-(2)
on solving (1) and (2)
d = 3 & a = -19
$S_{n}=\frac{n}{2}[2a+(n-1)d]$
$S_{12}=\frac{12}{2}[2(-19)+(12-1)(3)]$
$S_{12}=(6)[-38+33]$
$S_{12}=-30$
So the answer is option C.
4) Answer (A)
First term of AP = $a=-20$ and last term = $l=28$
Number of terms = $n=17$
Sum of AP = $\frac{n}{2}(a+l)$
= $\frac{17}{2}(-20+28)$
= $17 \times 4=68$
=> Ans – (A)
5) Answer (A)
$T_{3}$ = a + 2d = 13——-(1)
$T_{5}$ = a + 4d = 21——-(2)
on solving (1) AND (2)
d = 4 & a = 5
$T_{13}$ = a + 12d = 5 + 12(4) = 5 + 48 = 53
So the answer is option A.
6) Answer (C)
3 numbers are 43,91,183
largest number is 183
smallest number is 43
subtract smallest number from both the highest number
so 183 – 43 = 140
91 – 43 = 48
91 is smaller than 183, so subtract 91 from 183
183 – 91 = 92
now we have 3 numbers 140,48,92
so HCF of 140,48,92 = 4
thus the greatest number,that divides 43, 91 and 183 so as to leave the same remainder in each case, is 4
7) Answer (C)
8) Answer (D)
Subtract each number by 26
so we get
460 – 26 = 434
491 – 26 = 465
553 – 26 = 527
now taking HCF (434, 465 , 527) = 31 ( since 31 divides all the 3 numbers)
9) Answer (C)
10) Answer (D)
Arithmetic mean of 7, 5, 13, x and 9 = 10
=> $\frac{(7+5+13+x+9)}{5}=10$
=> $34+x=10 \times 5=50$
=> $x=50-34=16$
=> Ans – (D)
Get 200 SSC mocks for just Rs. 249. Enroll here
We hope this AP & GP questions for SSC Exam will be highly useful for your preparation.
