{"id":39439,"date":"2019-12-18T12:15:51","date_gmt":"2019-12-18T06:45:51","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=39439"},"modified":"2019-12-18T12:15:51","modified_gmt":"2019-12-18T06:45:51","slug":"progressions-questions-for-xat-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/progressions-questions-for-xat-set-2-pdf\/","title":{"rendered":"Progressions Questions for XAT Set-2 PDF"},"content":{"rendered":"

Progressions Questions for XAT Set-2 PDF<\/span><\/strong><\/span><\/h2>\n

Download important Progression Questions Set-2 for XAT\u00a0 Set-2 PDF based on previously asked questions in the XAT exam. Practice Progression Questions\u00a0 Set-2 PDF for the XAT exam.<\/p>\n

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Question 1:\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

Question 2:\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

Question 3:\u00a0<\/b>a,\u00a0 b,\u00a0 c,\u00a0 d and\u00a0 e are integers such that 1 \u2264 a < b < c < d < e. If a, b, c, d and e are geometric progression<\/span> and lcm (m , n) is the least common multiple of m and n, then the maximum value of $\\frac{1}{lcm(a,b)}+\\frac{1}{lcm(b,c)}+\\frac{1}{lcm(c,d)}+\\frac{1}{lcm(d,e)}$ is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a015\/16<\/p>\n

c)\u00a078\/81<\/p>\n

d)\u00a07\/8<\/p>\n

e)\u00a0None of these<\/p>\n

Question 4:\u00a0<\/b>If 2, a, b, c, d, e, f and 65 form an arithmetic progression, find out the value of \u2018e\u2019.<\/p>\n

a)\u00a048<\/p>\n

b)\u00a047<\/p>\n

c)\u00a041<\/p>\n

d)\u00a0None of the above<\/p>\n

Question 5:\u00a0<\/b>Suppose a, b and c are in Arithmetic Progression and $a^{2}, b^{2}$ and $c^{2}$ are in Geometric Progression. If $a<b<c$ and a+b+c=$\\frac{3}{2}$,, then the value of a=<\/p>\n

a)\u00a0$\\dfrac{1}{2\\sqrt{2}}$<\/p>\n

b)\u00a0$\\dfrac{1}{2\\sqrt{3}}$<\/p>\n

c)\u00a0$\\dfrac{1}{2} – \\dfrac{1}{\\sqrt{3}}$<\/p>\n

d)\u00a0$\\dfrac{1}{2} – \\dfrac{1}{\\sqrt{2}}$<\/p>\n

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Question 6:\u00a0<\/b>If three positive real numbers a, b and c (c > a) are in Harmonic Progression, then log (a + c) + log (a – 2b + c) is equal to:<\/p>\n

a)\u00a02 log (c – b)<\/p>\n

b)\u00a02 log (a – c)<\/p>\n

c)\u00a02 log (c – a)<\/p>\n

d)\u00a0log a + log b + log c<\/p>\n

Question 7:\u00a0<\/b>The interior angles of a polygon are in Arithmetic Progression. If the smallest angle is 120\u00b0 and common difference is 5\u00b0, then number of sides in the polygon is:<\/p>\n

a)\u00a07<\/p>\n

b)\u00a08<\/p>\n

c)\u00a09<\/p>\n

d)\u00a0None of the above<\/p>\n

Question 8:\u00a0<\/b>If the positive real numbers a, b and c are in Arithmetic Progression, such that abc = 4, then minimum possible value of b is:<\/p>\n

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

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

c)\u00a0$2^{\\frac{1}{3}}$<\/p>\n

d)\u00a0None of the above<\/p>\n

Question 9:\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

Question 10:\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

XAT Decision making practice questions<\/a><\/p>\n

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

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

The sum of the 3rd and 15th terms is a+2d+a+14d = 2a+16d
\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
\n2)\u00a0Answer\u00a0(A)<\/strong><\/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

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

Given that the numbers are in G.P.<\/p>\n

Let the common ratio be \u2018r\u2019, hence the series a,b,c,d,e can also be expressed as:<\/p>\n

$a , ar , ar^2 , ar^3 , ar^4$<\/p>\n

lcm(a,b) = lcm$(a,ar) = ar$<\/p>\n

lcm(b,c) = lcm$(ar,ar^2) = ar^2$<\/p>\n

lcm(c,d) = lcm$(ar^2,ar^3) = ar^3$<\/p>\n

lcm(d,e) = lcm$(ar^3,ar^4) = ar^4$<\/p>\n

$\\therefore \\frac{1}{lcm(a,b)}+\\frac{1}{lcm(b,c)}+\\frac{1}{lcm(c,d)}+\\frac{1}{lcm(d,e)}$<\/p>\n

= $\\frac{1}{ar} + \\frac{1}{ar^2} + \\frac{1}{ar^3} + \\frac{1}{ar^4}$<\/p>\n

= $\\frac{1}{a} (\\frac{1}{r} + \\frac{1}{r^2} + \\frac{1}{r^3} + \\frac{1}{r^4})$<\/p>\n

To get max value of this, \u2018a\u2019 and \u2018r\u2019 should be minimum.<\/p>\n

It is given that $1 \\leq a$ => Minimum value of \u2018a\u2019 = 1<\/p>\n

For the values in the series to be integers, the minimum common ratio, r = 2\u00a0\u00a0 ($r \\leq 1$ won\u2019t work here as it is an increasing GP)<\/p>\n

Substituting values of ‘a’ and ‘r’ in the expression, we get :<\/p>\n

Max value = $\\frac{1}{1} (\\frac{1}{2} + \\frac{1}{2^2} + \\frac{1}{2^3} + \\frac{1}{2^4})$<\/p>\n

= $\\frac{8 + 4 + 2 + 1}{16} = \\frac{15}{16}$<\/p>\n

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

Given that\u00a02, a, b, c, d, e, f and 65 are in an AP.<\/p>\n

65 = 2 + (8-1)d<\/p>\n

d = 9.<\/p>\n

Therefore, e = 2+(6-1)*9 = 2+45 = 47. Therefore, option B is the correct answer.<\/p>\n

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

Let us assume that the common difference of the A.P. is ‘d’.<\/p>\n

Then, we can say that a = b – d, c = b + d<\/p>\n

It is given that a + b + c = 3\/2. i.e. b = 1\/2.<\/p>\n

It is given that\u00a0$a^{2}, b^{2}$ and $c^{2}$ are in Geometric Progression. Hence, we can say that<\/p>\n

$b^4 = a^2*c^2$<\/p>\n

$b^4 = (b – d)^2*(b + d)^2$<\/p>\n

$b^4 = (b^2 – d^2)^2$<\/p>\n

$\\Rightarrow$ $(b^2+d^2-b^2)(b^2-d^2+b^2)=0$<\/p>\n

Therefore, $(b^2 – d^2 + b^2) = 0$. i.e. $d = \\dfrac{1}{\\sqrt{2}}$<\/p>\n

Hence, a = b – d = $\\dfrac{1}{2}$ – $\\dfrac{1}{\\sqrt{2}}$.<\/p>\n

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

It has been given that the terms $a, b,$ and $c$ are in harmonic progression.
\nTherefore, $\\frac{1}{b} – \\frac{1}{a} = \\frac{1}{c} – \\frac{1}{b}$
\n$\\frac{2}{b}$ = $\\frac{1}{a}$ + $\\frac{1}{c}$
\n$\\frac{2}{b} = \\frac{a+c}{ac}$
\n$b = \\frac{2ac}{(a+c)}$————–(1)
\nThe given expression is log $(a+c)$ + log $(a-2b+c)$.
\nlog $(a+c) + log (a – 2b + c)$ = log $((a+c)(a-2b + c))$
\nSubstituting (1), we get,
\nlog $(a+c)$ + log $(a – 2b + c)$ = log$((a+c)(a – \\frac{4ac}{(a+c)} +c))$
\n= log ($a^2 + ac – 4ac + c^2 + ac$)
\n= log $(a^2 + c^2 – 2ac)$
\n= log $(c-a)^2$ [Since c is greater than a]
\n= 2 log $(c-a)$
\nTherefore, option\u00a0C is the right answer.<\/p>\n

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

It has been given that the interior angles in a polygon are in an arithmetic progression.
\nWe know that the sum of all exterior angles of a polygon is 360\u00b0.
\nExterior angle = 180\u00b0 – interior angle.
\nSince we are subtracting the interior angles from a constant, the exterior angles will also be in an AP.
\nThe starting term of the AP formed by the exterior angles will be 180\u00b0-120\u00b0 = 60\u00b0 and the common difference will be -5\u00b0.<\/p>\n

Let the number of sides in the polygon be ‘n’.
\n=> The number of terms in the series will also be ‘n’.<\/p>\n

We know that the sum of an AP is equal to 0.5*n*(2a + (n-1)d), where ‘a’ is the starting term and ‘d’ is the common difference.
\n0.5*n*(2*60\u00b0 + (n-1)*(-5\u00b0)) = 360\u00b0
\n120$n$ – 5$n^2$ + 5$n$ = 720
\n5$n^2$ – 125$n$ + 720 = 0
\n$n^2$ – 25$n$ + 144 =0.
\n$(n-9)(n-16) = 0$<\/p>\n

Therefore, $n$ can be 9 or 16.
\nIf the number of sides is 16, then the largest external angle will be 60 – 15*5 = -15\u00b0. Therefore, we can eliminate this case.
\nThe number of sides in the polygon must be 9. Therefore, option C is the right answer.<\/p>\n

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

It has been given that a, b, and c are in an arithmetic progression.
\nLet a = x-p, b = x, and c = x+p
\nWe know that a, b, and c are real numbers.
\nTherefore, the arithmetic mean of a,b,c should be greater than or equal to the geometric mean.
\n$\\frac{a+b+c}{3} \\geq \\sqrt[3]{abc}$
\n$\\frac{a+b+c}{3} \\geq \\sqrt[3]{4}$
\n$\\frac{3x}{3}\\geq\\sqrt[3]{4}$<\/p>\n

$x\\geq\\sqrt[3]{4}$
\nWe know that $x$ = $b$.
\nTherefore,$b\\geq\\sqrt[3]{4}$or\u00a0$b\\geq 2^{\\frac{2}{3}}$
\nTherefore, option\u00a0B is the right answer.<\/p>\n

9)\u00a0Answer\u00a0(A)<\/strong><\/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

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

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We hope this Progression Questions Set-2 PDF for XAT with Solutions will be helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Progressions Questions for XAT Set-2 PDF Download important Progression Questions Set-2 for XAT\u00a0 Set-2 PDF based on previously asked questions in the XAT exam. Practice Progression Questions\u00a0 Set-2 PDF for the XAT exam. Question 1:\u00a0The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th […]<\/p>\n","protected":false},"author":42,"featured_media":39443,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3,3167,125,350,362,366],"tags":[3317,1425],"class_list":{"0":"post-39439","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"category-downloads-en","9":"category-featured","10":"category-iift","11":"category-snap","12":"category-xat","13":"tag-progression-questions-for-xat","14":"tag-xat"},"better_featured_image":{"id":39443,"alt_text":"Progressions Questions for XAT Set-2 PDF","caption":"Progressions Questions for XAT Set-2 PDF\n","description":"Progressions Questions for XAT Set-2 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