Progressions and Series is one of the most important topics in the Quantitative Ability section of CAT. It is an easy topic and so one must not avoid this topic. You can check out these Progressions and Series<\/span> CAT Previous year questions<\/a>. Practice a good number of questions on CAT Progressions and Series<\/strong> questions. In this article, we will look into some important Progressions and a Series of Questions for CAT. These are a good source for practice; If you want to practice these questions, you can download this CAT progression and Series Questions PDF 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 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 a)\u00a0u<\/p>\n b)\u00a0v<\/p>\n c)\u00a0w<\/p>\n d)\u00a0x<\/p>\n 2)\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 3:\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 3)\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 4:\u00a0<\/b>If $a_1 = 1$ and $a_{n+1} = 2a_n +5$, n=1,2,….,then $a_{100}$ is equal to:<\/p>\n a)\u00a0$(5*2^{99}-6)$<\/p>\n b)\u00a0$(5*2^{99}+6)$<\/p>\n c)\u00a0$(6*2^{99}+5)$<\/p>\n d)\u00a0$(6*2^{99}-5)$<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $a_2 = 2*1 + 5$ Question 5:\u00a0<\/b>What is the value of the following expression? a)\u00a09\/19<\/p>\n b)\u00a010\/19<\/p>\n c)\u00a010\/21<\/p>\n d)\u00a011\/21<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $(1\/(2^2-1))+(1\/(4^2-1))+(1\/(6^2-1))+…+(1\/(20^2-1)$ = 1\/[(2+1)*(2-1)] + 1\/[(4+1)*(4-1)] + … + 1\/[(20+1)*(20-1)]<\/p>\n = 1\/(1*3) + 1\/(3*5) + 1\/(5*7) + … + 1\/(19*21)<\/p>\n =1\/2 * ( 1\/1 – 1\/3 + 1\/3 – 1\/5 + 1\/5 – 1\/7 + … +1\/19 – 1\/21)<\/p>\n =1\/2 * (1 – 1\/21) = 10\/21<\/p>\n Question 6:\u00a0<\/b>The value of $\\frac{1}{1-x}+\\frac{1}{1+x}+\\frac{2}{1+x^2}+\\frac{4}{1+x^4}$<\/p>\n a)\u00a0$\\frac{8}{1-x^8}$<\/p>\n b)\u00a0$\\frac{4x}{1+x^2}$<\/p>\n c)\u00a0$\\frac{4}{1-x^6}$<\/p>\n d)\u00a0$\\frac{4}{1+x^4}$<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $\\frac{1}{1-x}+\\frac{1}{1+x}+\\frac{2}{1+x^2}+\\frac{4}{1+x^4}$ Question 7:\u00a0<\/b>N the set of natural numbers is partitioned into subsets $S_{1}$ = $(1)$, $S_{2}$ = $(2,3)$, $S_{3}$ =$(4,5,6)$, $S_{4}$ = $(7,8,9,10)$ and so on. The sum of the elements of the subset $S_{50}$ is<\/p>\n a)\u00a061250<\/p>\n b)\u00a065525<\/p>\n c)\u00a042455<\/p>\n d)\u00a062525<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n According to given question $S_{50}$ will have 50 terms Question 8:\u00a0<\/b>If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is<\/p>\n a)\u00a02:3<\/p>\n b)\u00a03:2<\/p>\n c)\u00a03:4<\/p>\n d)\u00a04:3<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The seventh term of an AP = a + 6d. Third term will be a\u00a0+ 2d and second term will be a\u00a0+ 16d. We are given that Question 9:\u00a0<\/b>Let $a_1$, $a_2$,………….,\u00a0 $a_{3n}$ be an arithmetic progression with $a_1$ = 3 and $a_{2}$ = 7. If $a_1$+ $a_{2}$ +…+ $a_{3n}$= 1830, then what is the smallest positive integer m such that m($a_1$+ $a_{2}$ +…+ $a_n$) > 1830?<\/p>\n a)\u00a08<\/p>\n b)\u00a09<\/p>\n c)\u00a010<\/p>\n d)\u00a011<\/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$a_3 = 2*(2 + 5) + 5 = 2^2 + 5*2 + 5$
\n$a_4 = 2^3 + 5*2^2 + 5*2 + 5$
\n…
\n$a_{100} = 2^{99} + 5*(2^{98} + 2^{97} + … + 1)$
\n$= 2^{99} + 5*1*(2^{99} – 1)\/(2-1) = 2^{99} + 5*2^{99} – 5 = 6*2^{99} – 5$<\/p>\n
\n$(1\/(2^2-1))+(1\/(4^2-1))+(1\/(6^2-1))+…+(1\/(20^2-1)$<\/p>\n
\nor $\\frac{2}{1-x^2}+\\frac{2}{1+x^2}+\\frac{4}{1+x^4}$
\nor\u00a0$\\frac{4}{1-x^4}+\\frac{4}{1+x^4}$
\nor\u00a0$\\frac{8}{1-x^8}$<\/p>\n
\nAnd its first term will be 50th number in the series 1,2,4,7,………$T_{50}$
\n$T_1 = 1$
\n$T_2 = 1+1$
\n$T_3 = 1+1+2$
\n$T_4 = 1+1+2+3$
\n$T_n = 1+(1+2+3+4+5….(n-1))$
\n= $1+\\frac{n(n-1)}{2}$
\nSo $T_{50} = 1+1225 = 1226$
\nHence $S_{50} = (1226,1227,1228,1229……..)$
\nAnd summation will be = $\\frac{50}{2} (2\\times 1226 + 49 \\times 1 ) = 62525$<\/p>\n
\n$ (a + 6d)^2 = (a + 2d)(a + 16d)$
\n=> $ a^2 $ + $36d^2$ +\u00a012ad = $ a^2 + 18ad + 32d^2 $
\n=> $4d^2 = 6ad$
\n=> $ d:a = 3:2$
\nWe have been asked about a:d. Hence, it would be 2:3<\/p>\n