{"id":212693,"date":"2022-07-11T17:58:36","date_gmt":"2022-07-11T12:28:36","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=212693"},"modified":"2022-07-11T17:58:36","modified_gmt":"2022-07-11T12:28:36","slug":"progression-and-series-questions-for-cat","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/progression-and-series-questions-for-cat\/","title":{"rendered":"Progression and Series Questions for CAT"},"content":{"rendered":"

Progression and Series<\/b> is an important topic in the CAT<\/strong> Quants<\/strong> (Algebra) section. <\/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> and tricks on how to answer questions<\/strong><\/span><\/a><\/span> on<\/strong><\/span> Progression and Series<\/strong><\/span><\/a>. You can check out these CAT Progression and Series<\/span> questions from the CAT Previous year papers<\/span><\/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

Enroll for CAT 2022 Online Course<\/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
\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

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

$a_2 = 2*1 + 5$
\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

Question 5:\u00a0<\/b>What is the value of the following expression?
\n$(1\/(2^2-1))+(1\/(4^2-1))+(1\/(6^2-1))+…+(1\/(20^2-1)$<\/p>\n

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

Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/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

$\\frac{1}{1-x}+\\frac{1}{1+x}+\\frac{2}{1+x^2}+\\frac{4}{1+x^4}$
\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

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
\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

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
\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

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

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

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

$a_{1}$ = 3 and\u00a0$a_{2}$ = 7. Hence, the common difference of the AP is 4.
\nWe have been given that the sum up to 3n terms of this AP is 1830. Hence, $1830 = \\frac{m}{2}[2*3 + (m – 1)*4$
\n=> 1830*2 = m(6 + 4m – 4)
\n=> 3660 = 2m + 4m$^2$
\n=> $2m^2 + m – 1830 = 0$
\n=> (m – 30)(2m + 61) = 0
\n=> m = 30 or m = -61\/2
\nSince m is the number of terms so m cannot be negative. Hence, must be 30
\nSo, 3n = 30
\nn = 10
\nSum of the first ’10’ terms of the given AP = 5*(6 + 9*4) = 42*5 = 210
\nm($a_1$+\u00a0$a_{2}$\u00a0+…+ $a_n$) > 1830
\n=> 210m >\u00a01830
\n=> m > 8.71
\nHence, smallest integral value of ‘m’ is 9.<\/p>\n

Question 10:\u00a0<\/b>Let $a_{1},a_{2},a_{3},a_{4},a_{5}$ be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with $2a_{3}$
\nIf the sum of the numbers in the new sequence is 450, then $a_{5}$ is<\/p>\n

10)\u00a0Answer:\u00a051<\/b><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Sum of the sequence of even numbers is $2a_{3} + (2a_{3} – 2) + (2a_{3} – 4)$ $+ (2a_{3} – 6) + (2a_{3} – 8) = 450$
\n=> $10a_{3} – 20 = 450$
\n=> $a_{3} = 47$
\nHence $a_{5} = 47 + 4 = 51$<\/p>\n

Question 11:\u00a0<\/b>An infinite geometric progression $a_1,a_2,…$ has the property that $a_n= 3(a_{n+1}+ a_{n+2} + …)$ for every n $\\geq$ 1. If the sum $a_1+a_2+a_3…+=32$, then $a_5$ is<\/p>\n

a)\u00a01\/32<\/p>\n

b)\u00a02\/32<\/p>\n

c)\u00a03\/32<\/p>\n

d)\u00a04\/32<\/p>\n

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

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Let the common ratio of the G.P. be r.
\nHence we have $a_n= 3(a_{n+1}+ a_{n+2} + …)$<\/p>\n

The sum up to infinity of GP is given by\u00a0$\\frac{a}{1-r}$ where a here is\u00a0$a_{n+1}$<\/p>\n

=> $a_n= 3(\\frac{a_{n+1}}{1-r})$
\n=> $a_n= 3(\\frac{a_{n}\\times r}{1-r})$
\n=> $ r = \\frac{1}{4}$
\nNow, $a_1+a_2+a_2…+=32$
\n=> $\\frac{a_1}{1-r} = 32$
\n=> $\\frac{a_1}{3\/4} = 32$
\n=> $a_1 = 24$<\/p>\n

$a_5 = a_1 \\times r^4$
\n$a_5 = 24 \\times (1\/4)^4 = \\frac{3}{32}$<\/p>\n

Question 12:\u00a0<\/b>If $a_{1}=\\frac{1}{2\\times5},a_{2}=\\frac{1}{5\\times8},a_{3}=\\frac{1}{8\\times11},…,$ then $a_{1}+a_{2}+a_{3}+…+a_{100}$ is<\/p>\n

a)\u00a0$\\frac{25}{151}$<\/p>\n

b)\u00a0$\\frac{1}{2}$<\/p>\n

c)\u00a0$\\frac{1}{4}$<\/p>\n

d)\u00a0$\\frac{111}{55}$<\/p>\n

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

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

$a_{100} = \\frac{1}{ (3\\times100 -1) \\times (3\\times100 + 2)}= \\frac{1}{ 299 \\times 302}$<\/p>\n

$\\frac{1}{2\\times5} = \\frac{1}{3} \\times (\\frac{1}{2} – \\frac{1}{5})$<\/p>\n

$\\frac{1}{5\\times8} = \\frac{1}{3} \\times (\\frac{1}{5} – \\frac{1}{8})$<\/p>\n

$\\frac{1}{8\\times11} = \\frac{1}{3} \\times (\\frac{1}{8} – \\frac{1}{11})$
\n….<\/p>\n

$\\frac{1}{299\\times302} = \\frac{1}{3} \\times (\\frac{1}{299} – \\frac{1}{302})$<\/p>\n

Hence $a_{1}+a_{2}+a_{3}+…+a_{100}$ = $\\frac{1}{3} \\times (\\frac{1}{2} – \\frac{1}{5})$ + $\\frac{1}{3} \\times (\\frac{1}{5} – \\frac{1}{8})$ + $\\frac{1}{3} \\times (\\frac{1}{8} – \\frac{1}{11})$ + … + $\\frac{1}{3} \\times (\\frac{1}{299} – \\frac{1}{302})$<\/p>\n

= $\\frac{1}{3} \\times (\\frac{1}{2} – \\frac{1}{302})$<\/p>\n

= $\\frac{25}{151}$<\/p>\n

Question 13:\u00a0<\/b>Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is<\/p>\n

a)\u00a0$\\frac{3}{6}$<\/p>\n

b)\u00a0$\\frac{1}{6}$<\/p>\n

c)\u00a0$\\frac{5}{2}$<\/p>\n

d)\u00a0$\\frac{3}{2}$<\/p>\n

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

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Let x = $a$, y = $ar$ and z = $ar^2$
\nIt is given that, 5x, 16y and 12z are in AP.
\nso, 5x + 12z = 32y
\nOn replacing the values of x, y and z, we get
\n$5a + 12ar^2 = 32ar$
\nor, $12r^2 – 32r + 5$ = 0
\nOn solving, $r$ = $\\frac{5}{2}$ or $\\frac{1}{6}$<\/p>\n

For $r$ =\u00a0$\\frac{1}{6}$, x < y < z is not satisfied.<\/p>\n

So,\u00a0$r$ =\u00a0$\\frac{5}{2}$<\/p>\n

Hence, option C is the correct answer.<\/p>\n

Question 14:\u00a0<\/b>Let $\\ a_{1},a_{2}…a_{52}\\ $ be positive integers such that $\\ a_{1}$ < $a_{2}$ < … <\u00a0$a_{52}\\ $. Suppose, their arithmetic mean is one less than arithmetic mean of $a_{2}$, $a_{3}$, ….$a_{52}$. If $a_{52}$= 100, then the largest possible value of $a_{1}$is<\/p>\n

a)\u00a048<\/p>\n

b)\u00a020<\/p>\n

c)\u00a023<\/p>\n

d)\u00a045<\/p>\n

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

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Let ‘x’ be the average of all 52\u00a0positive integers\u00a0$\\ a_{1},a_{2}…a_{52}\\ $.<\/p>\n

$a_{1}+a_{2}+a_{3}+…+a_{52}$ = 52x … (1)<\/p>\n

Therefore, average of\u00a0$a_{2}$, $a_{3}$, ….$a_{52}$ = x+1<\/p>\n

$a_{2}+a_{3}+a_{4}+…+a_{52}$ = 51(x+1)\u00a0… (2)<\/p>\n

From equation (1) and (2), we can say that<\/p>\n

$a_{1}+51(x+1)$ =\u00a052x<\/p>\n

$a_{1}$ = x – 51.<\/p>\n

We have to find out the largest possible value of\u00a0$a_{1}$.\u00a0$a_{1}$ will be maximum when ‘x’ is maximum.<\/p>\n

(x+1) is the average of terms\u00a0$a_{2}$, $a_{3}$, ….$a_{52}$. We know that\u00a0$a_{2}$ < $a_{3}$ < … <\u00a0$a_{52}\\ $ and\u00a0$a_{52}$ = 100.<\/p>\n

Therefore, (x+1) will be maximum when each term is maximum possible. If\u00a0$a_{52}$ = 100, then\u00a0$a_{52}$ = 99,\u00a0$a_{50}$ = 98 ends so on.<\/p>\n

$a_{2}$ = 100 + (51-1)*(-1) = 50.<\/p>\n

Hence, \u00a0$a_{2}+a_{3}+a_{4}+…+a_{52}$ = 50+51+…+99+100 =\u00a051(x+1)<\/p>\n

$\\Rightarrow$ $\\dfrac{51*(50+100)}{2} =\u00a051(x+1)$<\/p>\n

$\\Rightarrow$ $x =\u00a074$<\/p>\n

Therefore,\u00a0the\u00a0largest possible value of\u00a0$a_{1}$ = x – 51 = 74 – 51 = 23.<\/p>\n

Question 15:\u00a0<\/b>The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + …+ 95 x 99 is<\/p>\n

15)\u00a0Answer:\u00a080707<\/b><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

S = 7 x 11 + 11 x 15 + 15 x 19 + …+ 95 x 99<\/p>\n

Nth term of the series can be written as $T_{n} = (4n+3)*(4n+7)$<\/p>\n

Last term,\u00a0(4n+3) = 95 i.e. n = 23<\/p>\n

$\\sum_{n=1}^{n=23} (4n+3)*(4n+7)$<\/p>\n

$\\Rightarrow$ $\\sum_{n=1}^{n=23}16n^2+40n+21$<\/p>\n

$\\Rightarrow$\u00a0$16*\\dfrac{23*24*47}{6}+40*\\dfrac{23*24}{2}+21*23$<\/p>\n

$\\Rightarrow$\u00a0$80707$<\/p>\n

Enroll for CAT 2022 Complete Course <\/a><\/p><\/div>\n

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Download CAT Quant Formulas PDF<\/a><\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Progression and Series is an important topic in the CAT Quants (Algebra) section. If you find these questions a bit tough, make sure you solve more CAT Progression and Series questions. Learn all the important formulas and tricks on how to answer questions on Progression and Series. 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