Progressions Questions asked in SSC CGL

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Progressions Questions asked in SSC CGL
Progressions Questions asked in SSC CGL

Progressions Questions asked in SSC CGL:

Download SSC CGL Questions on Progressions with answers PDF based on previous papers very useful for SSC CGL exams. Top-10 Very Important History Questions for SSC Exams.

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Question 1: The first and last terms of an arithmetic progression are -32 and ­43. If the sum of the series is ­88, then it has how many terms?

a) 16

b) 15

c) 17

d) 14

Question 2: Find the value of p if 3x + p, x – 10 and -x + 16 are in arithmetic progression.

a) 16

b) 36

c) -16

d) -36

Question 3: The 4th term of an arithmetic progression is 15, 15th term is -29, find the 10th term?

a) -5

b) -13

c) -17

d) -9

Question 4: In an arithmetic progression if 13 is the 3rd term, ­47 is the 13th term, then ­30 is which term?

a) 9

b) 10

c) 7

d) 8

Question 5: The first and last terms of an arithmetic progression are 37 and ­-18. If the sum of the series is 114, then it has how many terms?

a) 13

b) 12

c) 14

d) 15

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Question 6: In an arithmetic progression, if 17 is the 3rd term, -25 is the 17th term, then -1 is which term?

a) 10

b) 11

c) 9

d) 12

Question 7: In an arithmetic progression, if 9 is the 5th term, -26 is the 12th term, then -6 is which term?

a) 11

b) 8

c) 10

d) 7

Question 8: Find the value of p, if 2x – 4, 4x + p and 6x – 12 are in arithmetic progression.

a) -9

b) -10

c) -11

d) -8

Question 9: Find the value of p if -3x – 11, x + p and 5x + 7 are in arithmetic progression.

a) 9

b) 2

c) -9

d) -2

Question 10: The first and last terms of an arithmetic progression are 29 and -49. If the sum of the series is -140, then it has how many terms?

a) 13

b) 14

c) 12

d) 11

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Answers & Solutions:

1) Answer (A)

First term of AP, $a=-32$ and last term, $l=43$

Let there be $n$ terms

Sum of AP = $\frac{n}{2}(a+l) = 88$

=> $\frac{n}{2}(-32+43)=88$

=> $\frac{11n}{2}=88$

=> $n=88 \times \frac{2}{11}$

=> $n=8 \times 2=16$

=> Ans – (A)

2) Answer (D)

Terms in arithmetic progression : $(3x + p) , (x – 10) , (-x + 16)$

=> Difference between first two terms is equal to the difference between last two terms

=> $(x – 10) – (3x + p) = (-x + 16) – (x – 10)$

=> $-2x -10 – p = -2x + 16 + 10$

=> $-p = 26 + 10 = 36$

=> $p = -36$

=> Ans – (D)

3) Answer (D)

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

4th term, $A_4 = a + (4 – 1) d = 15$

=> $a + 3d = 15$ —————–(i)

Similarly, 15th term, $A_{15} = a + 14d = -29$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(14d – 3d) = -29 – 15$

=> $d = \frac{-44}{11} = -4$

Substituting it in equation (i), => $a – 12 = 15$

=> $a = 15 + 12 = 27$

$\therefore$ 10th term, $A_{10} = a + (10 – 1)d$

= $27 + (9 \times -4) = 27 – 36 = -9$

=> Ans – (D)

4) Answer (D)

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

3rd term, $A_3 = a + (3 – 1) d = 13$

=> $a + 2d = 13$ —————–(i)

Similarly, 13th term, $A_{13} = a + 12d = 47$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(12d – 2d) = 47 – 13 = 34$

=> $d = \frac{34}{10} = 3.4$

Substituting it in equation (i), => $a + 2 \times 3.4 = 13$

=> $a = 13 – 6.8 = 6.2$

Let $n^{th}$ term = 30

=> $a + (n – 1) d = 30$

=> $6.2 + (n – 1) (3.4) = 30$

=> $(n – 1) (3.4) = 30 – 6.2 = 23.8$

=> $(n – 1) = \frac{23.8}{3.4} = 7$

=> $n = 7 + 1 = 8$

5) Answer (B)

In an arithmetic progression with first term, $a = 37$ , last term, $l = -18$

Let number of terms = $n$

$\therefore$ Sum of A.P. = $\frac{n}{2} (a + l) = 114$

=> $\frac{n}{2} (37 – 18) = 114$

=> $19n = 114 \times 2 = 228$

=> $n = \frac{228}{19} = 12$

=> Ans – (B)

6) Answer (C)

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

3rd term, $A_3 = a + (3 – 1) d = 17$

=> $a + 2d = 17$ —————–(i)

Similarly, 17th term, $A_{17} = a + 16d = -25$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(16d – 2d) = -25 – 17$

=> $d = \frac{-42}{14} = -3$

Substituting it in equation (i), => $a – 6 = 17$

=> $a = 17 + 6 = 23$

Let $n^{th}$ term = -1

=> $a + (n – 1) d = -1$

=> $23 + (n – 1) (-3) = -1$

=> $(n – 1) (-3) = -1 – 23 = -24$

=> $(n – 1) = \frac{-24}{-3} = 8$

=> $n = 8 + 1 = 9$

7) Answer (B)

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

5th term, $A_5 = a + (5 – 1) d = 9$

=> $a + 4d = 9$ —————–(i)

Similarly, 12th term, $A_{12} = a + 11d = -26$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(11d – 4d) = -26 – 9$

=> $d = \frac{-35}{7} = -5$

Substituting it in equation (i), => $a – 20 = 9$

=> $a = 9 + 20 = 29$

Let $n^{th}$ term = -6

=> $a + (n – 1) d = -6$

=> $29 + (n – 1) (-5) = -6$

=> $(n – 1) (-5) = -6 – 29 = -35$

=> $(n – 1) = \frac{-35}{-5} = 7$

=> $n = 7 + 1 = 8$

8) Answer (D)

Terms in arithmetic progression : $(2x – 4) , (4x + p) , (6x – 12)$

=> Difference between first two terms is equal to the difference between last two terms

=> $(4x + p) – (2x – 4) = (6x – 12) – (4x + p)$

=> $2x + p + 4 = 2x – 12 – p$

=> $2p = -12 – 4 = -16$

=> $p = \frac{-16}{2} = -8$

9) Answer (D)

Terms in arithmetic progression : $(-3x – 11) , (x + p) , (5x + 7)$

=> Difference between first two terms is equal to the difference between last two terms

=> $(x + p) – (-3x – 11) = (5x + 7) – (x + p)$

=> $4x + p + 11 = 4x + 7 – p$

=> $2p = 7 – 11 = -4$

=> $p = \frac{-4}{2} = -2$

=> Ans – (D)

10) Answer (B)

In an arithmetic progression with first term, $a = 29$ , last term, $l = -49$

Let number of terms = $n$

$\therefore$ Sum of A.P. = $\frac{n}{2} (a + l) = -140$

=> $\frac{n}{2} (29 – 49) = -140$

=> $\frac{-20n}{2} = -140$

=> $n = \frac{(-140) \times 2}{-20} = 7 \times 2$

=> $n=14$

=> Ans – (B)

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