Progressions Questions asked in SSC CGL:

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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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