Progression is an important topic in the Progression section of the SNAP Exam. You can also download this free Progression Questions PDF (with answers) for SNAP 2022 by Cracku. These questions will help you to practice and solve the Progression questions in the SNAP exam. Utilize this PDF practice set, <\/strong>which is one of the best sources for practising.<\/p>\n Download Progression Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\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 1)\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 2:\u00a0<\/b>Let the m-th and n-th terms of a geometric progression be $\\frac{3}{4}$ and 12. respectively, where $m < n$. If the common ratio of the progression is an integer r, then the smallest possible value of $r + n – m$ is<\/p>\n a)\u00a06<\/p>\n b)\u00a02<\/p>\n c)\u00a0-4<\/p>\n d)\u00a0-2<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the first term of the GP be “a” . Now from the question we can show that<\/p>\n $ar^{m-1}=\\frac{3}{4}$\u00a0 \u00a0\u00a0$ar^{n-1}=12$<\/p>\n Dividing both the equations we get\u00a0$r^{m-1-n+1}=\\frac{1}{16}\\ or\\ r^{m-n}=16^{-1\\ }or\\ r^{n-m}=16$<\/p>\n So for the minimum possible value we take Now give minimum possible value to “r” i.e -4 and n-m=2<\/p>\n Hence minimum possible value of r+n-m=-4+2=-2<\/p>\n Question 3:\u00a0<\/b>The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is<\/p>\n a)\u00a0-2<\/p>\n b)\u00a0-4<\/p>\n c)\u00a0-12<\/p>\n d)\u00a08<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the first term of the geometric progression be a and common ratio be r Question 4:\u00a0<\/b>What is the maximum sum of the terms in the arithmetic progression $25, 24\\frac{1}{2}, 24, …………..?$<\/p>\n a)\u00a0$637\\frac{1}{2}$<\/p>\n b)\u00a0625<\/p>\n c)\u00a0$662\\frac{1}{2}$<\/p>\n d)\u00a0650<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n It is a decreasing AP 25,24.5,24,23.5,…<\/p>\n The sum will be largest,\u00a0if nth term is zero then<\/p>\n 25 + (n-1)*(-.5) = 0<\/p>\n => n = 51<\/p>\n after 51 terms AP contains -ve terms<\/p>\n so, max sum will be obtained upto 50\/51th term<\/p>\n S = $\\frac{51}{2}$ *(25+0) = 637.5<\/p>\n Question 5:\u00a0<\/b>Consider an arithmetic progression of positive terms with the first term as $\\alpha$. Let $S_n$ denote the sum of the first n terms of this arithmetic progression and let $\\frac{S_m}{S_n} = \\frac{m^2}{n^2}$ for m \u2260 n. Then the $50^{th}$ term is<\/p>\n a)\u00a050 $\\alpha$<\/p>\n b)\u00a099\u00a0$\\alpha$<\/p>\n c)\u00a0100\u00a0$\\alpha$<\/p>\n d)\u00a0250\u00a0$\\alpha$<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{m^2}{n^2}$ =\u00a0$\\frac{\\frac{m*(2\\alpha+ (m-1)d)}{2}}{\\frac{n*(2\\alpha+ (n-1)d)}{2}}$<\/p>\n $\\frac{m}{n}$ = $\\frac{(2\\alpha+ (m-1)d)}{(2\\alpha+ (n-1)d)}$<\/p>\n $m(2\\alpha+ (n-1)d) = n(2\\alpha+ (m-1)d))$<\/p>\n $(n-m)d=(n-m)2\\alpha\\ $<\/p>\n d = 2$\\alpha$<\/p>\n $50^{th}$ term =\u00a0$\\alpha$+49*2$\\alpha$\u00a0= 99$\\alpha$<\/p>\n Question 6:\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 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let x = $a$, y = $ar$ and z = $ar^2$ 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 7:\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 7)\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. 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 => $a_n= 3(\\frac{a_{n+1}}{1-r})$ $a_5 = a_1 \\times r^4$ 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>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 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n It has been given that a, b, and c are in an arithmetic progression. $x\\geq\\sqrt[3]{4}$ Question 10:\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 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n It has been given that the interior angles in a polygon are in an arithmetic progression. Let the number of sides in the polygon be ‘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. Therefore, $n$ can be 9 or 16. Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 11:\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 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n It has been given that the terms $a, b,$ and $c$ are in harmonic progression. Question 12:\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 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/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 Question 13:\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 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/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 Question 14:\u00a0<\/b>In a list of 7 integers, one integer, denoted as x is unknown. The other six integers are 20, 4, 10, 4,8, and 4. If the mean, median, and mode of these seven integers are arranged in increasing order, they form an arithmetic progression. The sum of all possible values of x is<\/p>\n a)\u00a026<\/p>\n b)\u00a032<\/p>\n c)\u00a034<\/p>\n d)\u00a038<\/p>\n e)\u00a040<\/p>\n 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Integers = $4,4,4,8,10,20,x$<\/p>\n Clearly, irrespective of the value of $x$, Mode = $4$<\/p>\n Sum of above integers = $4 + 4 + 4 + 8 + 10 + 20 + x$<\/p>\n = $50 + x$<\/p>\n Mean = $\\frac{50 + x}{7}$<\/p>\n Case 1 : If $x < 4$<\/p>\n Median of $x,4,4,4,8,10,20$ = 4<\/p>\n Mode = 4<\/p>\n If these are in A.P. => Mean = 4<\/p>\n => $\\frac{50 + x}{7} = 4$<\/p>\n => $50 + x = 28$<\/p>\n => $x = 28 – 50 = -22$<\/p>\n => It is not possible<\/p>\n Case 2 : If $4 < x < 8$<\/p>\n Median of $4,4,4,x,8,10,20$ = $x$<\/p>\n Mode = 4<\/p>\n Mean = $\\frac{50 + x}{7}$<\/p>\n => $\\frac{54}{7} < Mean < \\frac{58}{7}$<\/p>\n As these are in AP => $x = 6$ and Mean = $8$<\/p>\n Case 3 : If $x > 8$<\/p>\n Mean = $\\frac{50 + x}{7} > \\frac{58}{7}$<\/p>\n Median of $4,4,4,8,x,10,20$ = $8$<\/p>\n Mode = $4$<\/p>\n As these are in AP, => Mean = $12$<\/p>\n => $\\frac{50 + x}{7} = 12$<\/p>\n => $50 + x = 84$<\/p>\n => $x = 84 – 50 = 34$<\/p>\n $\\therefore$ Sum of all possible values of x is = $6 + 34 = 40$<\/p>\n Question 15:\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 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/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 Question 16:\u00a0<\/b>If a, b, c, d, e and f are in arithmetic progression, then e-c is equal to<\/p>\n a)\u00a02(b-a)<\/p>\n b)\u00a0c-b<\/p>\n c)\u00a02(b-f)<\/p>\n d)\u00a02(d-b)<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n a, b, c, d, e, f are in AP => They can be written as a, a+x, a+2x, a+3x, a+4x and a+5x.<\/p>\n e – c = a + 4x – a – 2x = 2x<\/p>\n b – a = a + x – a = x<\/p>\n => e – c = 2(b – a)<\/p>\n Question 17:\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 17)\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 18:\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 18)\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)\u00a01<\/p>\n b)\u00a00<\/p>\n c)\u00a02<\/p>\n d)\u00a0-1<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the first term of the AP be $a$ and the common difference = $d$<\/p>\n 7th term = $A_7=a+6d$<\/p>\n 11th term = $A_{11}=a+10d$<\/p>\n According to ques,<\/p>\n => $7 \\times (a+6d)=11 \\times (a+10d)$<\/p>\n => $7a+42d=11a+110d$<\/p>\n => $11a-7a=42d-110d$<\/p>\n => $4a=-68d$<\/p>\n => $a=-17d$<\/p>\n => $a+17d = 0 = A_{18}$<\/p>\n => Ans – (B)<\/p>\n Question 20:\u00a0<\/b>What is the sum of the first 13 terms of an arithmetic progression if the first term is -10 and last term is 26?<\/p>\n a)\u00a0104<\/p>\n b)\u00a0140<\/p>\n c)\u00a084<\/p>\n d)\u00a098<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $S_{n}=\\frac{n}{2}[a+l]$<\/p>\n $S_{13}=\\frac{13}{2}[-10+26]$<\/p>\n $S_{13}=\\frac{13}{2}[16]=104$<\/p>\n So the answer is option A.<\/p>\n
\nNow as per given condition :
\na+ar= a(1+r) =12\u00a0 (1)
\nar^2+ar^3 = 48 (2)
\nwe get ar^2(1+r) =48\u00a0 (3)
\nDividing (1) and (3)
\nwe get :
\n1\/r^2 = 1\/4
\nwe get 1\/r = 1\/2 or -1\/2
\nbut since terms are alternatively positive and negative
\nwe get r = -2
\nSubstituting
\nwe get a(-1) =12
\nwe get a=-12<\/p>\n
\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
\nHence we have $a_n= 3(a_{n+1}+ a_{n+2} + …)$<\/p>\n
\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
\n$a_5 = 24 \\times (1\/4)^4 = \\frac{3}{32}$<\/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
\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
\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
\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
\n=> The number of terms in the series will also be ‘n’.<\/p>\n
\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
\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
\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
\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 19:\u00a0<\/b>If 7 times the seventh term of an Arithmetic Progression (AP) is equal to 11 times its eleventh term, then the 18th term of the AP will be<\/p>\n