Here you can download a free Number Series questions PDF with answers for IBPS PO and IBPS RRB PO 2022 by Cracku. These are some tricky questions in the IBPS PO and IBPS RRB PO 2022 exam that you need to find the Number Series for the given questions. These questions will help you to do practice and solve the Number Series questions in the IBPS PO and IBPS RRB PO exams. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Number Series MCQ<\/strong> PDF for IBPS PO and IBPS RRB PO 2022 for free.<\/p>\n Download Number Series Questions for IBPS RRB and PO Prelims<\/a><\/p>\n Download IBPS PO Previous Papers<\/a><\/p>\n Instructions<\/b><\/p>\n What will come in place of question mark (?) in the following series.<\/p>\n Question 1:\u00a0<\/b>62, ?, 514, 1032, 4130, 8264, 33058<\/p>\n a)\u00a0132<\/p>\n b)\u00a0136<\/p>\n c)\u00a0140<\/p>\n d)\u00a0128<\/p>\n e)\u00a0None of the above<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $62\\times2+4 = 124+4 = 128$ Here 128 will come in place of question mark. Question 2:\u00a0<\/b>180, 90, 135, ?, 1181.25, 5315.625<\/p>\n a)\u00a0332.5<\/p>\n b)\u00a0348.5<\/p>\n c)\u00a0359.5<\/p>\n d)\u00a0351.5<\/p>\n e)\u00a0337.5<\/p>\n 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n $180\\times0.5 = 90$ Here 337.5 will come in place of question mark. Question 3:\u00a0<\/b>326, ?, 295, 124, 268, 149, 245<\/p>\n a)\u00a0120<\/p>\n b)\u00a0135<\/p>\n c)\u00a095<\/p>\n d)\u00a080<\/p>\n e)\u00a0None of the above<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $326-21\\times11 = 326-231 = 95$ Here 95 will come in place of question mark. Question 4:\u00a0<\/b>6.2, 8.2, 12.7, 20.2, 30.7, ?<\/p>\n a)\u00a047.2<\/p>\n b)\u00a045.8<\/p>\n c)\u00a044.2<\/p>\n d)\u00a048.8<\/p>\n e)\u00a0None of the above<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $6.2+1.5\\times2 = 6.2+3 = 8.2$ Here 47.2 will come in place of question mark. Question 5:\u00a0<\/b>183, 1911, 580, 1580, 851, ?, 1020<\/p>\n a)\u00a0339<\/p>\n b)\u00a0351<\/p>\n c)\u00a01347<\/p>\n d)\u00a01363<\/p>\n e)\u00a0None of the above<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $183+12^{3} = 183+1728 = 1911$ Here 1363 will come in place of question mark. Instructions<\/b><\/p>\n Identify the wrong numbers in series I and series II and answer the question accordingly. Question 6:\u00a0<\/b>Find out the difference between the wrong numbers of series I and II.<\/p>\n a)\u00a0496.5<\/p>\n b)\u00a0475<\/p>\n c)\u00a0418.5<\/p>\n d)\u00a0362<\/p>\n e)\u00a0351<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n series I : series II : Difference between the wrong numbers of series I and II = 694.5-276 Instructions<\/b><\/p>\n Find out the missing number in each series and answer the questions as per the options.<\/p>\n Question 7:\u00a0<\/b>Series I : 3000, 1200, (A), 1152, 1843.2, 3686.4 a)\u00a01375<\/p>\n b)\u00a01445<\/p>\n c)\u00a01480<\/p>\n d)\u00a01465<\/p>\n e)\u00a01360<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Series I : Series II : Series III : value of (A+B+C) = 960+180+305 Question 8:\u00a0<\/b>Series I : (P), 9484, 16324, 21220, 24580, 26764 a)\u00a0R>Q>P<\/p>\n b)\u00a0Q>R>P<\/p>\n c)\u00a0R>P>Q<\/p>\n d)\u00a0P>Q>R<\/p>\n e)\u00a0P>Q=R<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Series I : Series II : Series III : R>P>Q Instructions<\/b><\/p>\n Study the following information carefully. J, K, L, M, 4125, 4414, __, Y<\/p>\n $J = P^{2}+Q$ Question 9:\u00a0<\/b>Which of the following statements is\/are true? a)\u00a0Only (i)<\/p>\n b)\u00a0Only (ii)<\/p>\n c)\u00a0Only (iii)<\/p>\n d)\u00a0Only (i) and (ii)<\/p>\n e)\u00a0Only (ii) and (iii)<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n (i) The value of Q is equal to the largest root of the equation $y^{2}-6y-3712 = 0$. Question 10:\u00a0<\/b>What is the value of \u2018Y\u2019?<\/p>\n a)\u00a04754<\/p>\n b)\u00a04736<\/p>\n c)\u00a04762<\/p>\n d)\u00a04708<\/p>\n e)\u00a0None of the above<\/p>\n 10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n (i) The value of Q is equal to the largest root of the equation $y^{2}-6y-3712 = 0$. Take Free IBPS PO Mock Tests<\/a><\/p>\n Instructions<\/b><\/p>\n Identify the missing numbers in Series I and Series II and answer question. Question 11:\u00a0<\/b>Which of the following statements are true about the missing numbers in the given series. a)\u00a0Only (i)<\/p>\n b)\u00a0Only (ii) and (iii)<\/p>\n c)\u00a0Only (i) and (ii)<\/p>\n d)\u00a0Only (ii)<\/p>\n e)\u00a0Only (iii)<\/p>\n 11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Series I Series II Hence, option e is the correct answer.<\/p>\n Instructions<\/b><\/p>\n Identify the wrong number in Series I and correct that number which will be the value of \u2018A\u2019. In Series II, find out the missing value in place question mark which will be the value of \u2018B\u2019. Question 12:\u00a0<\/b>Find out the difference between the values of \u2018A\u2019 and \u2018B\u2019.<\/p>\n a)\u00a01758.25<\/p>\n b)\u00a02712.25<\/p>\n c)\u00a01936.25<\/p>\n d)\u00a02374.25<\/p>\n e)\u00a0None of the above<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Series I Series II<\/b> difference between the values of \u2018A\u2019 and \u2018B\u2019 = 2412-37.75 Instructions<\/b><\/p>\n Find out the wrong number in the following series.<\/p>\n Question 13:\u00a0<\/b>3150, 2028, 1210, 648, 272, 100<\/p>\n a)\u00a0648<\/p>\n b)\u00a01210<\/p>\n c)\u00a0272<\/p>\n d)\u00a03150<\/p>\n e)\u00a0100<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $15^{3}-15^{2} = 3375-225 = 3150$ Here 272 is the wrong number. Question 14:\u00a0<\/b>464, 687, 1374, 3435, 10305, 36067.5<\/p>\n a)\u00a0687<\/p>\n b)\u00a010305<\/p>\n c)\u00a03435<\/p>\n d)\u00a0464<\/p>\n e)\u00a01374<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $458\\times1.5 = 687$ Here 464 is the wrong number. Question 15:\u00a0<\/b>1186, 1378, 1578, 1818, 2090, 2396<\/p>\n a)\u00a01818<\/p>\n b)\u00a02396<\/p>\n c)\u00a01378<\/p>\n d)\u00a02090<\/p>\n e)\u00a01578<\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $1186+13\\times14 = 1186+182 = 1368$ Here 1378 is the wrong number. Question 16:\u00a0<\/b>164, 180, -76, 1124, -2876, 7124, -13612<\/p>\n a)\u00a0-76<\/p>\n b)\u00a0180<\/p>\n c)\u00a0-2876<\/p>\n d)\u00a0164<\/p>\n e)\u00a01124<\/p>\n 16)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n $164+2^{4} = 164+16 = 180$ Here 1124 is the wrong number. Instructions<\/b><\/p>\n Find out the wrong number in the following series.<\/p>\n Question 17:\u00a0<\/b>312, 368, 259, 480, 32, 928<\/p>\n a)\u00a0928<\/p>\n b)\u00a0480<\/p>\n c)\u00a0368<\/p>\n d)\u00a032<\/p>\n e)\u00a0259<\/p>\n 17)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n $312+56=368$ Question 18:\u00a0<\/b>286, 288, 290, 296, 320, 440<\/p>\n a)\u00a0286<\/p>\n b)\u00a0320<\/p>\n c)\u00a0288<\/p>\n d)\u00a0296<\/p>\n e)\u00a0290<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern here is Question 19:\u00a0<\/b>928, 332, 285, 431, 1031, 4320<\/p>\n a)\u00a0431<\/p>\n b)\u00a0332<\/p>\n c)\u00a01031<\/p>\n d)\u00a0285<\/p>\n e)\u00a04320<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern here is Instructions<\/b><\/p>\n Find out the missing number in place of a question mark.<\/p>\n Question 20:\u00a0<\/b>180, 126, ?, 124.74, 162.162, 243.243<\/p>\n a)\u00a0115.4<\/p>\n b)\u00a0114.3<\/p>\n c)\u00a0114.5<\/p>\n d)\u00a0113.4<\/p>\n e)\u00a0112.3<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $180\\times0.7 = 126$ Here 113.4 is the missing number.
\n$128\\times4+2 = 512+2 = 514$
\n$514\\times2+4 = 1028+4 = 1032$
\n$1032\\times4+2 = 4128+2 = 4130$
\n$4130\\times2+4 = 8260+4 = 8264$
\n$8264\\times4+2 = 33056+2 = 33058$<\/p>\n
\nHence, option d is the correct answer.<\/p>\n
\n$90\\times1.5 = 135$
\n$135\\times2.5 = 337.5$
\n$337.5\\times3.5 = 1181.25$
\n$1181.25\\times4.5 = 5315.625$<\/p>\n
\nHence, option e is the correct answer.<\/p>\n
\n$95+20\\times10 = 95+200 = 295$
\n$295-19\\times9 = 295-171 = 124$
\n$124+18\\times8 = 124+144 = 268$
\n$268-17\\times7 = 268-119 = 149$
\n$149+16\\times6 = 149+96 = 245$<\/p>\n
\nHence, option c is the correct answer.<\/p>\n
\n$8.2+1.5\\times3 = 8.2+4.5 = 12.7$
\n$12.7+1.5\\times5 = 12.7+7.5 = 20.2$
\n$20.2+1.5\\times7 = 20.2+10.5 = 30.7$
\n$30.7+1.5\\times11 = 30.7+16.5 = 47.2$<\/p>\n
\nHence, option a is the correct answer.<\/p>\n
\n$1911-11^{3} = 1911-1331 = 580$
\n$580+10^{3} = 580+1000 = 1580$
\n$1580-9^{3} = 1580-729 = 851$
\n$851+8^{3} = 851+512 = 1363$
\n$1363-7^{3} = 1363-343 = 1020$<\/p>\n
\nHence, option d is the correct answer.<\/p>\n
\nseries I : 149, 66.5, 130.5, 45, 694.5, 4536, 74304
\nseries II : 78, 183, 276, 364, 436, 493<\/p>\n
\n$149\\times0.5-2^{3} = 74.5-8 = 66.5$
\n$66.5\\times1+4^{3} = 66.5+64 = 130.5$
\n$130.5\\times2-6^{3} = 261-216 = 45$
\n$45\\times4+8^{3} = 180+512 = 692$
\n$692\\times8-10^{3} = 5536-1000 = 4536$
\n$4536\\times16+12^{3} = 72576+1728 = 74304$
\nHere 694.5 is the wrong number in the series.<\/p>\n
\n$78+15\\times7 = 78+105 = 183$
\n$183+16\\times6 = 183+96 = 279$
\n$279+17\\times5 = 279+85 = 364$
\n$364+18\\times4 = 364+72 = 436$
\n$436+19\\times3 = 436+57 = 493$
\nHere 276 is the wrong number in the series.<\/p>\n
\n= 418.5
\nHence, option c is the correct answer.<\/p>\n
\nSeries II : (B), 960, 3765, 11385, 22665, 22785
\nSeries III : 1220, 309, 150.5, 154.5, (C), 1224, 9788
\nWhat is the value of (A+B+C)?<\/p>\n
\n$3000\\times0.4 = 1200$
\n$1200\\times0.8 = 960$
\n$960\\times1.2 = 1152$
\n$1152\\times1.6 = 1843.2$
\n$1843.2\\times2 = 3686.4$<\/p>\n
\n$180\\times5+60 = 900+60 = 960$
\n$960\\times4-75 = 3840-75 = 3765$
\n$3765\\times3+90 = 11295+90 = 11385$
\n$11385\\times2-105 = 22770-105 = 22665$
\n$22665\\times1+120 = 22665+120 = 22785$<\/p>\n
\n$1220\\times0.25+4 = 305+4 = 309$
\n$309\\times0.5-4 = 154.5-4 = 150.5$
\n$150.5\\times1+4 = 150.5+4 = 154.5$
\n$154.5\\times2-4 = 309-4 = 305$
\n$305\\times4+4 = 1220+4 = 1224$
\n$1224\\times8-4 = 9792-4 = 9788$<\/p>\n
\n= 1445
\nHence, option b is the correct answer.<\/p>\n
\nSeries II : 24, (Q), 960, 3000, 7560, 16464
\nSeries III : 3265920, 322560, 35280, 4320, (R), 96<\/p>\n
\n$244 + 21^{3} – 21 = 244 + 9261 – 21 = 9484$
\n$9484 + 19^{3} – 19 = 9484 + 6859 – 19 = 16324$
\n$16324 + 17^{3} – 17 = 16324 + 4913 – 17 = 21220$
\n$21220 + 15^{3} – 15 = 21220 + 3375 – 15 = 24580$
\n$24580 + 13^{3} – 13 = 24580 + 2197 – 13 = 26764$<\/p>\n
\n$2^{5}-2^{3} = 32-8 = 24$
\n$3^{5}-3^{3} = 243-27 = 216$
\n$4^{5}-4^{3} = 1024-64 = 960$
\n$5^{5}-5^{3} = 3125-125 = 3000$
\n$6^{5}-6^{3} = 7776-216 = 7560$
\n$7^{5}-7^{3} = 16807-343 = 16464$<\/p>\n
\n10!-9! = 3628800-362880 = 3265920
\n9!-8! = 362880-40320 = 322560
\n8!-7! = 40320-5040 = 35280
\n7!-6! = 5040-720 = 4320
\n6!-5! = 720-120 = 600
\n5!-4! = 120-24 = 96<\/p>\n
\n600>244>216
\nHence, option c is the correct answer.<\/p>\n
\nA series is given and certain conditions are given below for it. Observe the conditions carefully and answer the questions.<\/p>\n
\n$K = (T-1)^{2}+J$
\n$L = S^{2}+K$
\n
\n<\/b>Note ::
\n<\/b>(i) The value of Q is equal to the largest root of the equation $y^{2}-6y-3712 = 0$.
\n(ii) The difference between S and Q is 35.
\n(iii) LCM of P and T is 1184 where P is a prime number.
\n(iv) The value of S is not the multiple of 3.<\/p>\n
\n(i) The value of M is the multiple of 7.
\n(ii) The difference between the values of P and T is five.
\n(iii) The immediate next term after Y in the given series has 91 as a factor.<\/p>\n
\n$y^{2}-(64-58)y-3712 = 0$
\n$y^{2}-64y+58y-3712 = 0$
\ny(y-64)+58(y-64) = 0
\n(y-64) (y+58) = 0
\ny = -58, 64
\nlargest root of the equation = Q = 64
\n<\/strong>(ii) The difference between S and Q is 35.
\nS-64 = 35 \u21d2 S = 35+64 = 99
\n64-S = 35 \u21d2 S = 64-35 = 29
\n(iv) The value of S is not the multiple of 3.
\nS = 29<\/strong> (99 is the multiple of 3. So it will not be possible.)
\n(iii) LCM of P and T is 1184 where P is a prime number.
\nPT = $37\\times32$
\nSo P = 37 and T = 32
\n<\/strong>$J = P^{2}+Q = 37^{2}+64 = 1369+64 = 1433$
\n$K = (T-1)^{2}+J = (32-1)^{2}+1433 = 31^{2}+1433 = 961+1433 = 2394$
\n$L = S^{2}+K = 29^{2}+2394 = 841+2394 = 3235$
\nFrom these three numbers, now we can get the pattern from which we can obtain the value of Y.
\nM = $23^{2}+3235 = 529+3235 = 3764$
\n$19^{2}+3764 = 361+3764 = 4125$
\n$17^{2}+4125 = 289+4125 = 4414$
\n$13^{2}+4414 = 169+4414 = 4583$
\nY = $11^{2}+4583 = 121+4583 = 4704$
\n$7^{2}+4704 = 49+4704 = 4753$
\nHence, option b is the correct answer.<\/p>\n
\n$y^{2}-(64-58)y-3712 = 0$
\n$y^{2}-64y+58y-3712 = 0$
\ny(y-64)+58(y-64) = 0
\n(y-64) (y+58) = 0
\ny = -58, 64
\nlargest root of the equation = Q = 64
\n<\/strong>(ii) The difference between S and Q is 35.
\nS-64 = 35 \u21d2 S = 35+64 = 99
\n64-S = 35 \u21d2 S = 64-35 = 29
\n(iv) The value of S is not the multiple of 3.
\nS = 29<\/strong> (99 is the multiple of 3. So it will not be possible.)
\n(iii) LCM of P and T is 1184 where P is a prime number.
\nPT = $37\\times32$
\nSo P = 37 and T = 32
\n<\/strong>$J = P^{2}+Q = 37^{2}+64 = 1369+64 = 1433$
\n$K = (T-1)^{2}+J = (32-1)^{2}+1433 = 31^{2}+1433 = 961+1433 = 2394$
\n$L = S^{2}+K = 29^{2}+2394 = 841+2394 = 3235$
\nFrom these three numbers, now we can get the pattern from which we can obtain the value of Y.
\nM = $23^{2}+3235 = 529+3235 = 3764$
\n$19^{2}+3764 = 361+3764 = 4125$
\n$17^{2}+4125 = 289+4125 = 4414$
\n$13^{2}+4414 = 169+4414 = 4583$
\nY = $11^{2}+4583 = 121+4583 = 4704$
\nHence, option e is the correct answer.<\/p>\n
\nSeries I :: 469, 3629269, 3266389, 3306709, 3301669, ?, 3302269
\nSeries II :: 109, 125, 206, 831, 3232, 17873, ?<\/p>\n
\n(i) In Series II, the digits on the unit and tens place of the missing number are the same.
\n(ii) The sum of the digits of the missing number in Series I is 26.
\n(iii) The missing number in Series II is the multiple of three.<\/p>\n
\n<\/b>469 + 10! = 469 + 3628800 = 3629269
\n3629269 – 9! = 3629269 – 362880 = 3266389
\n3266389 + 8! = 3266389 + 40320 = 3306709
\n3306709 – 7! = 3306709 – 5040 = 3301669
\n3301669 + 6! = 3301669 + 720 = 3302389
\n3302389 – 5! = 3302389 – 120 = 3302269
\nHere 3302389 is the missing number in the given series.<\/p>\n
\n<\/b>$109+2^{4} = 109+16 = 125$
\n$125+3^{4} = 125+81 = 206$
\n$206+5^{4} = 206+625 = 831$
\n$831+7^{4} = 831+2401 = 3232$
\n$3232+11^{4} = 3232+14641 = 17873$
\n$17873+13^{4} = 17873+28561 = 46434$
\nHere 46434 is the missing number in the given series.<\/p>\n
\nSeries I :: 526, 275.5, 39.75, 101.75, 167.5, 686,\u00a05484
\nSeries II :: 1248, 1591, 1898, 2171, ?, 2623
\nNote :: both the series are following a different pattern.
\n<\/strong><\/p>\n
\n<\/b>$526\\times0.25 + 144 = 131.5+144 = 275.5$
\n$275.5\\times0.5 – 100 = 137.75-100 = 37.75$
\n$37.75\\times1 + 64 = 37.75+64 = 101.75$
\n$101.75\\times2 – 36 = 203.5-36 = 167.5$
\n$167.5\\times4 + 16 = 670+16 = 686$
\n$686\\times8 – 4 = 5488-4 = 5484$
\nHere 37.75 should come instead of 39.75 to make the series correct. So A = 37.75<\/p>\n
\n$1248+19^{2}-18 = 1248+361-18 = 1591$
\n$1591+18^{2}-17 = 1591+324-17 = 1898$
\n$1898+17^{2}-16 = 1898+289-16 = 2171$
\n$2171+16^{2}-15 = 2171+256-15 = 2412$
\n$2412+15^{2}-14 = 2412+225-14 = 2623$
\nThe missing number is 2412.<\/p>\n
\n= 2374.25
\nHence, option d is the correct answer.<\/p>\n
\n$13^{3}-13^{2} = 2197-169 = 2028$
\n$11^{3}-11^{2} = 1331-121 = 1210$
\n$9^{3}-9^{2} = 729-81 = 648$
\n$7^{3}-7^{2} = 343-49 = 294$
\n$5^{3}-5^{2} = 125-25 = 100$<\/p>\n
\nHence, option c is the correct answer.<\/p>\n
\n$687\\times2 = 1374$
\n$1374\\times2.5 = 3435$
\n$3435\\times3 = 10305$
\n$10305\\times3.5 = 36067.5$<\/p>\n
\nHence, option d is the correct answer.<\/p>\n
\n$1368+14\\times15 = 1368+210 = 1578$
\n$1578+15\\times16 = 1578+240 = 1818$
\n$1818+16\\times17 = 1818+272 = 2090$
\n$2090+17\\times18 = 2090+306 = 2396$<\/p>\n
\nHence, option c is the correct answer.<\/p>\n
\n$180-4^{4} = 180-256 = -76$
\n$-76+6^{4} = -76+1296 = 1220$
\n$1220-8^{4} = 1220-4096 = -2876$
\n$-2876+10^{4} = -2876+10000 = 7124$
\n$7124-12^{4} = 7124-20736 = -13612$<\/p>\n
\nHence, option e is the correct answer.<\/p>\n
\n$368-112=256$
\n$256+224=480$
\n$480-448=32$
\n$32+896=928$
\n$\\therefore$ The wrong number in the series is 259
\nHence, the correct answer is Option E<\/p>\n
\n$287+1!=287+1=288$
\n$288+2!=288+2=290$
\n$290+3!=290+6=296$
\n$296+4!=296+24=320$
\n$320+5!=320+120=440$
\n$\\therefore$ The wrong number in the series is 286
\nHence, the correct answer is Option A<\/p>\n
\n$928\\times0.25+100=232+100=332$
\n$332\\times0.5+121=166+121=287$
\n$287\\times1+144=287+144=431$
\n$431\\times2+169=862+169=1031$
\n$1031\\times4+196=4124+196=4320$
\n$\\therefore$ The wrong number in the series is 285
\nHence, the correct answer is Option D<\/p>\n
\n$126\\times0.9 = 113.4$
\n$113.4\\times1.1 = 124.74$
\n$124.74\\times1.3 = 162.162$
\n$162.162\\times1.5 = 243.243$<\/p>\n
\nHence, option d is the correct answer.<\/p>\n