{"id":40233,"date":"2020-01-29T17:25:09","date_gmt":"2020-01-29T11:55:09","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40233"},"modified":"2020-01-29T17:25:09","modified_gmt":"2020-01-29T11:55:09","slug":"progressions-questions-asked-in-ssc-cgl","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/progressions-questions-asked-in-ssc-cgl\/","title":{"rendered":"Progressions Questions asked in SSC CGL"},"content":{"rendered":"

Progressions Questions asked in SSC CGL:<\/span><\/strong><\/h1>\n

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.<\/p>\n

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Question 1:\u00a0<\/b>The first and last terms of an arithmetic progression are -32 and \u00ad43. If the sum of the series is \u00ad88, then it has how many terms?<\/p>\n

a)\u00a016<\/p>\n

b)\u00a015<\/p>\n

c)\u00a017<\/p>\n

d)\u00a014<\/p>\n

Question 2:\u00a0<\/b>Find the value of p if 3x + p, x – 10 and -x + 16 are in arithmetic progression.<\/p>\n

a)\u00a016<\/p>\n

b)\u00a036<\/p>\n

c)\u00a0-16<\/p>\n

d)\u00a0-36<\/p>\n

Question 3:\u00a0<\/b>The 4th term of an arithmetic progression is 15, 15th term is -29, \ufb01nd the 10th term?<\/p>\n

a)\u00a0-5<\/p>\n

b)\u00a0-13<\/p>\n

c)\u00a0-17<\/p>\n

d)\u00a0-9<\/p>\n

Question 4:\u00a0<\/b>In an arithmetic progression if 13 is the 3rd term, \u00ad47 is the 13th term, then \u00ad30 is which term?<\/p>\n

a)\u00a09<\/p>\n

b)\u00a010<\/p>\n

c)\u00a07<\/p>\n

d)\u00a08<\/p>\n

Question 5:\u00a0<\/b>The first and last terms of an arithmetic progression are 37 and \u00ad-18. If the sum of the series is 114, then it has how many terms?<\/p>\n

a)\u00a013<\/p>\n

b)\u00a012<\/p>\n

c)\u00a014<\/p>\n

d)\u00a015<\/p>\n

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Question 6:\u00a0<\/b>In an arithmetic progression, if 17 is the 3rd term, -25 is the 17th term, then -1 is which term?<\/p>\n

a)\u00a010<\/p>\n

b)\u00a011<\/p>\n

c)\u00a09<\/p>\n

d)\u00a012<\/p>\n

Question 7:\u00a0<\/b>In an arithmetic progression, if 9 is the 5th term, -26 is the 12th term, then -6 is which term?<\/p>\n

a)\u00a011<\/p>\n

b)\u00a08<\/p>\n

c)\u00a010<\/p>\n

d)\u00a07<\/p>\n

Question 8:\u00a0<\/b>Find the value of p, if 2x – 4, 4x + p and 6x – 12 are in arithmetic progression.<\/p>\n

a)\u00a0-9<\/p>\n

b)\u00a0-10<\/p>\n

c)\u00a0-11<\/p>\n

d)\u00a0-8<\/p>\n

Question 9:\u00a0<\/b>Find the value of p if -3x – 11, x + p and 5x + 7 are in arithmetic progression.<\/p>\n

a)\u00a09<\/p>\n

b)\u00a02<\/p>\n

c)\u00a0-9<\/p>\n

d)\u00a0-2<\/p>\n

Question 10:\u00a0<\/b>The \ufb01rst 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?<\/p>\n

a)\u00a013<\/p>\n

b)\u00a014<\/p>\n

c)\u00a012<\/p>\n

d)\u00a011<\/p>\n

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Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

First term of AP, $a=-32$ and last term,\u00a0$l=43$<\/p>\n

Let there be $n$ terms<\/p>\n

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

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

=> $\\frac{11n}{2}=88$<\/p>\n

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

=> $n=8 \\times 2=16$<\/p>\n

=> Ans – (A)<\/p>\n

2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Terms in arithmetic progression\u00a0: $(3x + p) , (x – 10) , (-x + 16)$<\/p>\n

=> Difference between first two terms is equal to the difference between last two terms<\/p>\n

=> $(x – 10) – (3x + p) = (-x + 16) – (x – 10)$<\/p>\n

=> $-2x -10 – p = -2x + 16 + 10$<\/p>\n

=> $-p = 26 + 10 = 36$<\/p>\n

=> $p = -36$<\/p>\n

=> Ans – (D)<\/p>\n

3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

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.<\/p>\n

4th term, $A_4 = a + (4 – 1) d = 15$<\/p>\n

=> $a + 3d = 15$ —————–(i)<\/p>\n

Similarly, 15th term, $A_{15} = a + 14d = -29$ ——————(ii)<\/p>\n

Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n

=> $(14d – 3d) = -29 – 15$<\/p>\n

=> $d = \\frac{-44}{11} = -4$<\/p>\n

Substituting it in equation (i), => $a – 12 = 15$<\/p>\n

=> $a = 15 + 12 = 27$<\/p>\n

$\\therefore$ 10th term, $A_{10} = a + (10 – 1)d$<\/p>\n

= $27 + (9 \\times -4) = 27 – 36 = -9$<\/p>\n

=> Ans – (D)<\/p>\n

4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

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.<\/p>\n

3rd term, $A_3 = a + (3 – 1) d = 13$<\/p>\n

=> $a + 2d = 13$ —————–(i)<\/p>\n

Similarly, 13th term, $A_{13} = a + 12d = 47$ ——————(ii)<\/p>\n

Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n

=> $(12d – 2d) = 47 – 13 = 34$<\/p>\n

=> $d = \\frac{34}{10} = 3.4$<\/p>\n

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

=> $a = 13 – 6.8 = 6.2$<\/p>\n

Let $n^{th}$ term = 30<\/p>\n

=> $a + (n – 1) d = 30$<\/p>\n

=> $6.2 + (n – 1) (3.4) = 30$<\/p>\n

=> $(n – 1) (3.4) = 30 – 6.2 = 23.8$<\/p>\n

=> $(n – 1) = \\frac{23.8}{3.4} = 7$<\/p>\n

=> $n = 7 + 1 = 8$<\/p>\n

5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

In an arithmetic progression with first term, $a = 37$ , last term, $l = -18$<\/p>\n

Let number of terms = $n$<\/p>\n

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

=> $\\frac{n}{2} (37 – 18) = 114$<\/p>\n

=> $19n = 114 \\times 2 = 228$<\/p>\n

=> $n = \\frac{228}{19} = 12$<\/p>\n

=> Ans – (B)<\/p>\n

6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

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.<\/p>\n

3rd term, $A_3 = a + (3 – 1) d = 17$<\/p>\n

=> $a + 2d = 17$ —————–(i)<\/p>\n

Similarly, 17th term, $A_{17} = a + 16d = -25$ ——————(ii)<\/p>\n

Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n

=> $(16d – 2d) = -25 – 17$<\/p>\n

=> $d = \\frac{-42}{14} = -3$<\/p>\n

Substituting it in equation (i), => $a – 6 = 17$<\/p>\n

=> $a = 17 + 6 = 23$<\/p>\n

Let $n^{th}$ term = -1<\/p>\n

=> $a + (n – 1) d = -1$<\/p>\n

=> $23 + (n – 1) (-3) = -1$<\/p>\n

=> $(n – 1) (-3) = -1 – 23 = -24$<\/p>\n

=> $(n – 1) = \\frac{-24}{-3} = 8$<\/p>\n

=> $n = 8 + 1 = 9$<\/p>\n

7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

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.<\/p>\n

5th term, $A_5 = a + (5 – 1) d = 9$<\/p>\n

=> $a + 4d = 9$ —————–(i)<\/p>\n

Similarly, 12th term, $A_{12} = a + 11d = -26$ ——————(ii)<\/p>\n

Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n

=> $(11d – 4d) = -26 – 9$<\/p>\n

=> $d = \\frac{-35}{7} = -5$<\/p>\n

Substituting it in equation (i), => $a – 20 = 9$<\/p>\n

=> $a = 9 + 20 = 29$<\/p>\n

Let $n^{th}$ term = -6<\/p>\n

=> $a + (n – 1) d = -6$<\/p>\n

=> $29 + (n – 1) (-5) = -6$<\/p>\n

=> $(n – 1) (-5) = -6 – 29 = -35$<\/p>\n

=> $(n – 1) = \\frac{-35}{-5} = 7$<\/p>\n

=> $n = 7 + 1 = 8$<\/p>\n

8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Terms in arithmetic progression\u00a0: $(2x – 4) , (4x + p) , (6x – 12)$<\/p>\n

=> Difference between first two terms is equal to the difference between last two terms<\/p>\n

=> $(4x + p) – (2x – 4) = (6x – 12) – (4x + p)$<\/p>\n

=> $2x + p + 4 = 2x – 12 – p$<\/p>\n

=> $2p = -12 – 4 = -16$<\/p>\n

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

9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Terms in arithmetic progression\u00a0: $(-3x – 11) , (x + p) , (5x + 7)$<\/p>\n

=> Difference between first two terms is equal to the difference between last two terms<\/p>\n

=> $(x + p) – (-3x – 11) = (5x + 7) – (x + p)$<\/p>\n

=> $4x + p + 11 = 4x + 7 – p$<\/p>\n

=> $2p = 7 – 11 = -4$<\/p>\n

=> $p = \\frac{-4}{2} = -2$<\/p>\n

=> Ans – (D)<\/p>\n

10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

In an arithmetic progression with first term, $a = 29$ , last term, $l = -49$<\/p>\n

Let number of terms = $n$<\/p>\n

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

=> $\\frac{n}{2} (29 – 49) = -140$<\/p>\n

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

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

=> $n=14$<\/p>\n

=> Ans – (B)<\/p>\n

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We hope this Progressions Questions for SSC CGL Exam preparation is so helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

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. Take a free mock test for SSC CGL Download SSC CGL Previous Papers PDF More SSC CGL Important Questions and Answers PDF […]<\/p>\n","protected":false},"author":21,"featured_media":40246,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,9,504,378,1493,1459,1611,1741,1268,1441,1],"tags":[3551,462],"class_list":{"0":"post-40233","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-ssc-cpo","13":"category-ssc-gd","14":"category-ssc-je","15":"category-ssc-mts","16":"category-ssc-stenographer","17":"category-stenographer","18":"category-uncategorized","19":"tag-progressions-questions","20":"tag-ssc-cgl"},"better_featured_image":{"id":40246,"alt_text":"Progressions Questions 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