Progression and Series<\/b> is an important topics in the CAT<\/strong> Quant<\/strong>\u00a0section. <\/span>If you find these questions a bit tough, make sure you solve more CAT Progression and Series questions. Learn all the important formulas<\/strong><\/span><\/a> and tricks on how to answer questions<\/span><\/strong><\/a><\/span> on<\/span> Progression and Series<\/span><\/strong><\/a>. You can check out these CAT progressions and series <\/span>questions from the CAT Previous year papers<\/strong><\/a>.<\/span> Practice a good number of sums in the CAT Progression and Series<\/strong> so that you can answer these questions with ease in the exam. In this post, we will look into some important Progression and Series Questions for CAT quants. These are a good source of practice for CAT 2022 preparation; If you want to practice these questions, you can download this Important CAT Progression and Series Questions<\/strong> and Answers PDF<\/strong> along with the video solutions below, which is completely Free.<\/p>\n Download Progression and Series Questions for CAT<\/a><\/p>\n CAT Online Coaching<\/a><\/p>\n Question 1:\u00a0<\/b>The number of common terms in the two sequences 17, 21, 25,\u2026, 417 and 16, 21, 26,\u2026, 466 is<\/p>\n a)\u00a078<\/p>\n b)\u00a019<\/p>\n c)\u00a020<\/p>\n d)\u00a077<\/p>\n e)\u00a022<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The terms of the first sequence are of the form 4p + 13<\/p>\n The terms of the second sequence are of the form 5q + 11<\/p>\n If a term is common to both the sequences, it is of the form 4p+13 and 5q+11<\/p>\n or 4p = 5q -2. LHS = 4p is always even, so, q is also even.<\/p>\n or 2p = 5r – 1 where q = 2r.<\/p>\n Notice that LHS is again even, hence r should be odd. Let r = 2m+1 for some m.<\/p>\n Hence, p = 5m + 2.<\/p>\n So, the number = 4p+13 = 20m + 21.<\/p>\n Hence, all numbers of the form 20m + 21 will be the common terms. i.e 21,41,61,…,401 = 20.<\/p>\n Question 2:\u00a0<\/b>The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?<\/p>\n a)\u00a01st<\/p>\n b)\u00a09th<\/p>\n c)\u00a012th<\/p>\n d)\u00a0None of the above<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The sum of the 3rd and 15th terms is a+2d+a+14d = 2a+16d a)\u00a0u<\/p>\n b)\u00a0v<\/p>\n c)\u00a0w<\/p>\n d)\u00a0x<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n 1, 2, 3, 4,….n such that the sum is greater than 288 Question 4:\u00a0<\/b>If the product of n positive real numbers is unity, then their sum is necessarily<\/p>\n a)\u00a0a multiple of n<\/p>\n b)\u00a0equal to n + 1\/n<\/p>\n c)\u00a0never less than n<\/p>\n d)\u00a0a positive integer<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the numbers be $a_1,a_2….a_n.$<\/p>\n Since the numbers are positive,<\/p>\n $AM\\geq GM$<\/p>\n $\\frac{a_1+a_2+a_3….+a_n}{n}\\geq (a_1*a_2….*a_n)^{1\/n}$<\/p>\n $a_1+a_2+a_3….+a_n \\geq n$<\/p>\n Question 5:\u00a0<\/b>If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms?<\/p>\n a)\u00a00<\/p>\n b)\u00a0-1<\/p>\n c)\u00a01<\/p>\n d)\u00a0Not unique<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Sum of the first 11 terms = 11\/2 ( 2a+10d)<\/p>\n Sum of the first 19 terms = 19\/2 (2a+18d)<\/p>\n => 22a+110d = 38a+342d => 16a = -232d<\/p>\n => 2a = -232\/8 d = -29d<\/p>\n Sum of the first 30 terms = 15(2a+29d) = 0<\/p>\n Question 6:\u00a0<\/b>Consider the sequence of numbers $a_1, a_2, a_3$……. to infinity where $a_1 = 81.33$ and $a_2 = -19$ and $a_j = a_{j-1} – a_{j-2}$ for $j > 3$. What is the sum of the first 6002 terms of this sequence?<\/p>\n a)\u00a0-100.33<\/p>\n b)\u00a0-30.00<\/p>\n c)\u00a062.33<\/p>\n d)\u00a0119.33<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n According to given conditions the terms are 81.33, -19, -100.33, -81.33, 19, 100.33, 81.33,-19,.. Hence the series repeats after every 6 terms . Also summation of these 6 terms is 0. Hence summation is 60002 terms will we sum of first 2 terms which is 62.33.<\/p>\n Question 7:\u00a0<\/b>Consider a sequence where the $n^{th}$ term, $t_n = n\/(n+2), n =1, 2, ….$ The value of $t_3 * t_4 * t_5 * \u2026..* t_{53}$ equals.<\/p>\n a)\u00a02\/495<\/p>\n b)\u00a02\/477<\/p>\n c)\u00a012\/55<\/p>\n d)\u00a01\/1485<\/p>\n e)\u00a01\/2970<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n substituting 3,4…53 in the\u00a0given function, we get Multiplying the values, we get\u00a0$\\frac{3}{5}*\\frac{4}{6}*\\frac{5}{7}*….\\frac{52}{54}*\\frac{53}{55} $ which ultimately after cancellations give $\\frac{3*4}{54*55}=\\frac{2}{495}$<\/p>\n Question 8:\u00a0<\/b>A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?<\/p>\n a)\u00a03<\/p>\n b)\u00a04<\/p>\n c)\u00a05<\/p>\n d)\u00a06<\/p>\n e)\u00a07<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let x be in the front row.<\/p>\n So no. of children in next rows will be x-3,x-6,x-9,x-12,x-15,x-18,x-21….<\/p>\n Suppose there are 6 rows, then the sum is equal to x + x-3 + x-6 + x-9 + x-12 + x-15 = 6x – 45<\/p>\n This sum is equal to 630.<\/p>\n => 6x – 45 = 630 => 6x = 585<\/p>\n Here, x is not an integer.<\/p>\n Hence, there cannot be 6 rows.<\/p>\n Question 9:\u00a0<\/b>For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of 7th and 6th terms of this sequence is 517, what is the 10th term of this sequence?<\/p>\n a)\u00a0147<\/p>\n b)\u00a076<\/p>\n c)\u00a0123<\/p>\n d)\u00a0Cannot be determined<\/p>\n
\nThe sum of the 6th, 11th and 13th terms is a+5d+a+10d+a+12d = 3a+27d
\nSince the two are equal, 2a+16d = 3a+27d => a+11d = 0
\nSo, the 12th term is 0
\nQuestion 3:\u00a0<\/b>The 288th term of the series a,b,b,c,c,c,d,d,d,d,e,e,e,e,e,f,f,f,f,f,f\u2026 is<\/p>\n
\nIf n = 24, n(n+1)\/2 = 12*25 = 300
\nSo, n = 24, i.e. the 24th letter in the alphabet is the letter at position 288 in the series
\n\u200bSo, answer = x<\/p>\n
\n$t_3 = \\frac{3}{5}$
\n$t_4 = \\frac{4}{6}$
\n$t_5 = \\frac{5}{7}$
\n$t_6 = \\frac{6}{8}$<\/p>\n