# AP GP Questions for SSC CGL

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## 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.

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

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 $\sqrt{5}$,$\sqrt{4}$,$\sqrt{2}$,$\sqrt{3}$ is

a) $\sqrt{4}$

b) $\sqrt{3}$

c) $\sqrt{5}$

d) $\sqrt{2}$

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

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)

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)

$T_{3}$ = 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.

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)

$T_{3}$ = 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.

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

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)

=> $\frac{(7+5+13+x+9)}{5}=10$
=> $34+x=10 \times 5=50$
=> $x=50-34=16$