This post will look at the important Progressions and Series of questions in the CAT Quant section. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download this CAT Progressions and Series Questions PDF (most important) along with the detailed solutions below, which is completely Free.<\/p>\n
Download Progressions and series (quants) questions for cat<\/a><\/p>\n Enroll for CAT 2023 Crash Course<\/a><\/p>\n Question 1:\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 1)\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 2:\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 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $a_2 = 2*1 + 5$ Question 3:\u00a0<\/b>A child was asked to add first few natural numbers (i.e. 1 + 2 + 3 + \u2026) so long his patience permitted. As he stopped, he gave the sum as 575. When the teacher declared the result wrong, the child discovered he had missed one number in the sequence during addition. The number he missed was<\/p>\n a)\u00a0less than 10<\/p>\n b)\u00a010<\/p>\n c)\u00a015<\/p>\n d)\u00a0more than 15<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n If the child adds all the numbers from 1 to 34, the sum of the numbers would be 1+2+3+…+34 = 34*35\/2 = 595 Question 4:\u00a0<\/b>A square is drawn by joining the midpoints of the sides of a given square. A third square is drawn inside the second square in the same way and this process is continued indefinitely. If a side of the first square is 8 cm, the sum of the areas of all the squares such formed (in sq.cm.)is<\/p>\n a)\u00a0128<\/p>\n b)\u00a0120<\/p>\n c)\u00a096<\/p>\n d)\u00a0None of these<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Side of first square = 8cm.<\/p>\n Side of second square made by joining mid-points of first square = $\\frac{8}{\\sqrt2}$<\/p>\n Similarly side of third square =\u00a0\u00a0$\\frac{8}{\\sqrt2\\times\\sqrt2}$ and so on.<\/p>\n Now summation of areas will be $8^2+(\\frac{8}{\\sqrt2})^2+(\\frac{8}{\\sqrt2\\times\\sqrt2})^2$ ………<\/p>\n or $8^2 ( 1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}…..)$<\/p>\n or $64 \\times (\\frac{1}{1-\\frac{1}{2}})$ (As we know sum of an infinite G.P. is $\\frac{a}{1-r}$ where a is first term and r is common ratio)<\/p>\n or 128 sq. cm.<\/p>\n Question 5:\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 5)\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 Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 6:\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 6)\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 7:\u00a0<\/b>Let $a_1, a_2, …$ be integers such that a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a010<\/p>\n d)\u00a0-1<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $a_1 – a_2 + a_3 – a_4 + …. + (-1)^{n – 1} a_n = n$<\/p>\n It is clear from the above equation that when n is odd, the co-efficient of a is positive otherwise negative.<\/p>\n $a_1 – a_2 = 2$<\/p>\n $a_1 = a_2 + 2$<\/p>\n $a_1 – a_2 + a_3 = 3$<\/p>\n On substituting the value of $a_1$ in the above equation, we get<\/p>\n $a_3$ = 1<\/p>\n $a_1 – a_2 + a_3 – a_4 = 4$<\/p>\n On substituting the values of $a_1, a_3$ in the above equation, we get<\/p>\n $a_4$ = -1<\/p>\n $a_1 – a_2 + a_3 – a_4 +a_5 = 5$<\/p>\n On substituting the values of $a_1, a_3, a_4$ in the above equation, we get<\/p>\n $a_5$ = 1<\/p>\n So we can conclude that $a_3, a_5, a_7….a_{n+1}$ = 1 and\u00a0$a_2, a_4, a_6….a_{2n}$ = -1<\/p>\n Now we have to find the value of\u00a0$a_{51} + a_{52} + …. + a_{1023}$<\/p>\n Number of terms = 1023=51+(n-1)1<\/p>\n n=973<\/p>\n There will be 486 even and 487 odd terms, so the value of\u00a0$a_{51} + a_{52} + …. + a_{1023}$ = 486*-1+487*1=1<\/p>\n Question 8:\u00a0<\/b>If $a_1 + a_2 + a_3 + …. + a_n = 3(2^{n + 1} – 2)$, for every $n \\geq 1$, then $a_{11}$ equals<\/p>\n 8)\u00a0Answer:\u00a06144<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n 11th term of series =\u00a0$a_{11}$ = Sum of 11 terms – Sum of 10 terms = $3(2^{11 + 1} – 2)$-3$(2^{10 + 1} – 2)$<\/p>\n = 3$(2^{12} – 2-2^{11} +2)$=3$(2^{11})(2-1)$= 3*$2^{11}$ = 6144<\/p>\n Question 9:\u00a0<\/b>Three positive integers x, y and z are in arithmetic progression. If $y-x>2$ and $xyz=5(x+y+z)$, then z-x equals<\/p>\n a)\u00a08<\/p>\n b)\u00a012<\/p>\n c)\u00a014<\/p>\n d)\u00a010<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Given x, y, z are three terms in an arithmetic progression.<\/p>\n Considering x = a, y = a+d, z = a+2*d.<\/p>\n Using the given equation x*y*z = 5*(x+y+z)<\/p>\n a*(a+d)*(a+2*d) = 5*(a+a+d+a+2*d)<\/p>\n =a*(a+d)*(a+2*d)\u00a0 = 5*(3*a+3*d) = 15*(a+d).<\/p>\n = a*(a+2*d) = 15.<\/p>\n Since all x, y, z are positive integers and y-x > 2. a, a+d, a+2*d are integers.<\/p>\n The common difference is positive and greater than 2.<\/p>\n Among the different possibilities are : (a=1, a+2d = 5), (a, =3, a+2d = 5), (a = 5, a+2d = 3), (a=15, a+2d = 1)<\/p>\n Hence the only possible case satisfying the condition is :<\/p>\n a = 1, a+2*d = 15.<\/p>\n x = 1, z = 15.<\/p>\n z-x = 14.<\/p>\n Question 10:\u00a0<\/b>Consider the arithmetic progression 3, 7, 11, … and let $A_n$ denote the sum of the first n terms of this progression.\u00a0Then the value of $\\frac{1}{25} \\sum_{n=1}^{25} A_{n}$ is<\/p>\n a)\u00a0455<\/p>\n b)\u00a0442<\/p>\n c)\u00a0415<\/p>\n d)\u00a0404<\/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
\nSince the child got the sum as 575, he would have missed the number 20.<\/p>\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2016\/01\/25\/11409.png\")
<\/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
\n$a_1 – a_2 + a_3 – a_4 + …. + (-1)^{n – 1} a_n = n,$ for all $n \\geq 1.$
\nThen $a_{51} + a_{52} + …. + a_{1023}$ equals<\/p>\n