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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 551. |
Simplest form of 368/496., 473/645, 1095/1168 |
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Answer» 368/496=23/31473/645=11/431095/1168=15/16 I don know what the answer is |
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| 552. |
3/2x+5/3y=7, 9x-10y=14 are consistent or inconsistent |
| Answer» consistent solution | |
| 553. |
What is reminder theorem |
| Answer» It states that the remainder of the division of a polynomial by a linear polynomial is equal to In particular, is a divisor of if and only if a property known as the factor theorem | |
| 554. |
If the zeroa of the polynomial x3 - 3x2 + x + 1 are a-b,a,a+b , find the value of (a + b) |
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| 555. |
If the point (-2,-1),(1,0),(x,3)and (1,y) form a parallelogram, find the value of x and y. |
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| 556. |
Chaptar=3X+3y=62x-3y=12 |
| Answer» Both eq. Is sameAnd its has infinity many sol. | |
| 557. |
1.1 Question 4,3,5 |
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Answer» NCERT Solutions for Class 10 Maths Exercise 1.13. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?Ans.\xa0We have to find the HCF (616, 32) to find the maximum number of columns in which they can march.To find the HCF, we can use Euclid’s algorithm.The HCF (616, 32) is 8.Therefore, they can march in 8 columns each. NCERT Solutions for Class 10 Maths Exercise 1.14. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m\xa0or 3m\xa0+ 1 for some integer\xa0m. [Hint: Let\xa0x\xa0be any positive integer then it is of the form 3q, 3q\xa0+ 1 or 3q\xa0+\xa02. Now square each of these and show that they can be rewritten in the form 3m\xa0or 3m +\xa01.]Ans.\xa0Let\xa0a\xa0be any positive integer and\xa0b\xa0= 3.Then\xa0a\xa0= 3q\xa0+\xa0r\xa0for some integer\xa0q\xa0≥ 0And\xa0r\xa0= 0, 1, 2 because 0 ≤\xa0r\xa0< 3Therefore,\xa0a\xa0= 3q\xa0or 3q\xa0+ 1 or 3q\xa0+ 2Or,Where\xa0\xa0are some positive integers.Hence, it can be said that the square of any positive integer is either of the form 3m\xa0or 3m\xa0+ 1. NCERT Solutions for Class 10 Maths Exercise 1.15. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m\xa0+ 1 or 9m +\xa08.Ans.\xa0Let\xa0a\xa0be any positive integer and\xa0b\xa0= 3a =\xa03q\xa0+\xa0r, where\xa0q\xa0≥ 0 and 0 ≤\xa0r\xa0< 3\\ a = 3q or 3q + 1 or 3q + 2Therefore, every number can be represented as these three forms.We have three cases.Case 1: When\xa0a = 3q,Where\xa0m\xa0is an integer such that\xa0m\xa0=3q3Case 2: When\xa0a\xa0= 3q\xa0+ 1,Where\xa0m\xa0is an integer such that\xa0Case 3: When\xa0a\xa0= 3q\xa0+ 2,Where\xa0m\xa0is an integer such that\xa0Therefore, the cube of any positive integer is of the form 9m, 9m\xa0+ 1, or 9m +\xa08. |
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| 558. |
The 7th and 12th terms of an AP are 46 and 71 respectively. Find the nth term of the AP |
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Answer» Sorry I think I did a mistake there should be a+(7-1)d=46=a+6d=46..1eq.,a+(12-1)d=71=a+11d=71.....2eq.Now if u will substract these two equation u will get the answer.. a+6(n-1)=46,a+11(n-1)=71Now substract these two equation,I hope you will get the answer... 6+(n-1)5 not sure about the answer. |
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| 559. |
Pair of linear equation in two variable |
| Answer» An\xa0equation\xa0that can be of the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a\xa0linear equation in two variables\xa0x and y. | |
| 560. |
HAPPY INDEPENDENCE DAY GUYS ????????????? |
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Answer» Happy independence day Happy Independence Day ???? |
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| 561. |
Pls tell all the formulas used in polynomials chapter |
| Answer» ????????????????????????????´◔‿ゝ◔`)━☞〈(•ˇ‿ˇ•)-→←(>▽<)ﾉ☜ (↼_↼)☜ (↼_↼)←(*꒪ヮ꒪*)(☞ ͡° ͜ʖ ͡°)☞(☉｡☉)!→(☞^o^) ☞(☞ﾟヮﾟ)☞(☉｡☉)!→(☞ﾟヮﾟ)☞(☉｡☉)!→(☞ﾟヮﾟ)☞☜ (↼_↼)(☞ﾟヮﾟ)☞☜ (↼_↼)(☞^o^) ☞⟵(๑¯◡¯๑)｡.ﾟ+ ⟵(｡･ω･)(╭☞•́⍛•̀)╭☞⟵(๑¯◡¯๑)(╭☞•́⍛•̀)╭☞⟵(๑¯◡¯๑)｡.ﾟ+ ⟵(｡･ω･)☜ (↼_↼)☜ (↼_↼) | |
| 562. |
Find the sum of first 51 terms of an A.P.whose second and third terms are 14 and 18 respectively |
| Answer» \xa0given,a2=14\xa0a3=18\xa0d=a3−a2=4\xa0a1=a2−d\xa0a1 =14-4a=10Sn=n/2(2a+(n-1)d)\xa0Sn=51/2(2*10+(51-1)4)\xa0Sn=5610 | |
| 563. |
What is consistent and in consistent in maths 3 chapter |
| Answer» If the lines intersect at a point, then that point gives the unique solution of the two eqution.In this case,the pair of equations is consistent.& If the lines are parallel,then the pair of equations has no solution.In this case,the pair of equations is inconsistent. | |
| 564. |
what is the answer of 2345-1234? |
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Answer» 1111 is the answer prithvi. 1111 1111 1111 1111 |
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| 565. |
Show that x=-bc/ad is a solution of the quardatic equation ad*2(ax/b+2c/d)x+bc*2=0 |
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| 566. |
Solve the quadratic equation 3xsq.-243=0 |
| Answer» 3x^2-243=0X^2-81=0X^2-9^2=0(a^2-b^2)=(a-b)(a+b)(X-9)(x+9)=0x-9=0,x+9=0x=9,x=-9 | |
| 567. |
Find the value of a and b for which x=3/4 and x=-2 are roots of the equation ax*2 + bx - 6=0. |
| Answer» Answer:The value of a and b are 4 and 5 resp.Step-by-step explanation:we have\xa0to find the value of a and b.The value of a and b are 4 and 5 resp. | |
| 568. |
If a and b are zeroes of polynomial x2+2x+1,then 1/a+1/b is? |
| Answer» 1/a +1/ b = a+ b / ab= - b/ a/ c/ a=- b/ c=-2/1 | |
| 569. |
If graph of y=x*x-1 is drawn at which of the points the graph will intersect x-axis. |
| Answer» At(1,0) | |
| 570. |
Find the zeros of x2-2x-8 |
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Answer» x2 - 2x - 8=x2 - (4 - 2)x - 8=x2 - 4x + 2x - 8=x(x - 4) + 2(x - 4)=(x - 4)(x + 2)=0 (x - 4)=0 or (x + 2)=0 x=4 x = -2So zeros are 4 and -2 x2-2x-8= x2\xa0- 4x + 2x - 8= x( x - 4) + 2 (x - 4)= ( x + 2) ( x - 4) x2 - 2x - 8 = 0= x2 - 4x + 2x - 8 = 0= x(x - 4) + 2(x - 4) = 0= (x + 2) (x - 4) = 0So zeros are -2 and 4 x2-2x-8= x2 - 4x + 2x - 8= x( x - 4) + 2 (x - 4)= ( x + 2) ( x - 4) |
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| 571. |
What is the cube of 5 |
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Answer» See 8f we ant to remove the cube of any no. then cube means what the multiplication of no.thre timesfore.eg-5×5×5=125 because 5×5=25 and25×5=125 125 Don\'t you know that 225 225 |
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| 572. |
Find the mid point of line segment joining points A(1,2) and B(4,6) |
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Answer» Sheetal has enough money to buy 5kg mangoes at the rate of RS 18 per kg how much quantity of mangoes she can buy in the same many of the price increase of to RS 20 per kg 1/5, 4 |
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| 573. |
Chapter=32x-3y=7(a+b) x-(a+b-3) y=4a+b |
| Answer» Value of a, b = -5, -1 derived from the system of linear equation given which has infinite number of solution.Solution:For an equation to have infinite solution, the equations should be same as for now let us equate the equations and then solve it2x - 3y=7 (a+b)x-(a+b-3)y = 4a+bEquate the value of ax + by = cn in both cases, after this we get:a+b=2na+b-3=-3n and 4a+b=7n. Solving these we get the value of n = -3.Putting the value of n, we get a+b=-6a+b-3=18 4a+b=-21Solving the above equation, we get:a=-6-b-4(6+b)+b=-21b= -1 Putting the value of b in a = -6-ba = -5.Therefore, the value of a & b = -5, -1 | |
| 574. |
Show that x=5, y=2 is a solution of the equation 2x+3y=16 & x-2y =1 |
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Answer» Put x = 5 and y = 2\xa02x+3y=162 (5) + 3 (2) = 10 + 6 = 16\xa0x-2y =15 - 2 (2) = 5 - 4 = 1Hence proved 2(5)+3(2)=10+6=16=R.H.SAnd5-2(2)=5-4=1=R.H.SHENCE PROOF. |
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| 575. |
A quadrilateral ABCD is drawn to circumscribe a circle prove that AB + CD is equals to ad + Bc |
| Answer» Since, tangents drawn from an exterior point to a circle are equal in length.∴ AP = AS ...(i)BP = BQ ...(ii)CR = CQ ...(iii)and DR = DS ...(iv)Adding (i), (ii), (iii) and (iv), we getAP + BP + CR + DR = AS + BQ + CQ + DS⇒ (AP + BP) + (CR + DR)= (AS + DS) + (BQ + CQ)⇒ AB + CD = AD + BCHence, AB + CD = BC + DA. | |
| 576. |
Muje math ka test dena h |
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Answer» Yes 1.proof that root is irrational |
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| 577. |
Prove thaat underroot 5is iirational |
| Answer» Let us assume that √5 is a rational number.Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0⇒√5=p/qOn squaring both the sides we get,⇒5=p²/q²⇒5q²=p² —————–(i)p²/5= q²So 5 divides pp is a multiple of 5⇒p=5m⇒p²=25m² ————-(ii)From equations (i) and (ii), we get,5q²=25m²⇒q²=5m²⇒q² is a multiple of 5⇒q is a multiple of 5Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number√5 is an irrational number | |
| 578. |
Find the higest common factor of the following by complete factorisation(1) 48,56,72 |
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Answer» 8 On |
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| 579. |
How do ancient Egyptians would write the fraction 6/29 |
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| 580. |
How we can find the value of √289 ? |
| Answer» Factorise it as a product of primes and you will find root 289 equals to √17×17 it is square root so you will make pair of two so root 289 equals to 17 | |
| 581. |
How many three digit number are divisible by 7? |
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Answer» The first 3-digit number which is divisible by 7 is 105The last 3-digit number which is divisible by 7 is 994The list of 3-digit numbers divisible by 7 are105, 112, 119,…..994 which forms an A.PConsider a formulaT(n) = a + (n – 1)dWherea = 105d = 7T(n) = 994994 = 105 + (n – 1)7889 = 7n – 77n = 896n = 128∴ There are 128 3-digits number which are divisible by 7. AP 105,112........994A=105 d=7 an=994n=? An= a+(n-1) DN=128There are 128 no. |
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| 582. |
Which term of the AP: 3,15,27,39,.....will be 132 more than its 54th term?? |
| Answer» let the required term be nth term54th term = a +(n-1)d\xa0= 3 + 53x12 = 3 + 636 = 639therefore 132 + 639 = 771 will be the nth term.771 = 3 + (n-1)12768/12 +1 = ntherefore n = 64\xa0+1 = 65therefore the 65th term will be 132 more than the 54th term | |
| 583. |
Can anyone share me their timetable?? I am confused? |
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Answer» Sumaila g ne muhe kha k phle apne textbook k chp count kro 145 chp ha aur baad ma daily ka ek chp complete kro jan me khtm ho jyga syllabus if we follow that . This is really helpful for me Thanks? It\'s your new time for your lock down period first I know that your online classes are going on whatever your time schedule is started from 12.00pm to 3.00pm is your first study session after one hour is your next study session which is 4.00pm to 5.30pm let\'s take a break and your third study session at 9.00pm to 11.00pm and your last thing is for last half hour you should revice that what you are study in your hole day ????✍️✍️???????? |
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| 584. |
In an equilateral triangle of side 3^3 cm find the length of the altitude |
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| 585. |
explain why 7*11*13*+13 and 7*6*5*4*3*2*1+5 |
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Answer» {FUNDAMENTAL THEOREM OF ARITHMETIC:Every composite number can be expressed as a product of primes,and this factorisation is unique,apart from the order in which the prime factors occur...}(I)=(7×11×13)+13=13(7×11×+1)=13×2×3×13 It can be represent in the form of prime factors...... Therefore,It is a composite number...(II)=7×6×5×4×3×2×1+5=5(7×6×4×3×2×1+1)=5×1009 It can be represent in the form of prime factors..... Therefore,It is a composite number..........Hope it helps uh Nd thank uh :) Use calci to solve it Use calci to solve it Use calculator |
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| 586. |
(x-a)/(x-b) + (x-b)/(x-a)= a/b+ b/a Solve this quadratic equation by factorization. |
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| 587. |
Find distance between points (a-b,a+b),(-a-b,a-b) |
| Answer» Your answer is √a²+b² | |
| 588. |
What is the Solution of equations9x + 7y = 55.....eq(1)7x + 9y = 57.....eq(2) |
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Answer» X=871/288,y=127/32 This is right and? X=3 Y=4 |
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| 589. |
Root 25 ka gunanafal gyaat kijiye |
| Answer» 5 | |
| 590. |
What is current acid |
| Answer» Both the current ratio, also known as the working capital ratio, and acid-test ratio measure a company\'s short-term ability to generate enough cash to pay off all debts should they become due at once. | |
| 591. |
Probability of hitting a ball is 7/5 then what is the probability of not hitting a ball |
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Answer» Tq sis How can it possible dear coz in probability there is numerator is smaller than denominator and the denominator shows the chances of things happen.... |
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| 592. |
What is the square root of 4 ? |
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Answer» 2 2 2 2 2 |
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| 593. |
-52+6 |
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Answer» I was just checking u all -46 -46....by the way are you really in 10th standard? -46 Shame |
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| 594. |
Rs Aggarwal solution rs Aggarwal solution |
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Answer» Are you talking about rd sharma? ?? |
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| 595. |
Find the number which is divisible by 11 |
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Answer» Find the square root of the following number by using their ones and tens digit ?(b) 1444 ,(c)3025 (d)4761 Test for divisibility by 11. Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number.So, for instance, 2728 has alternating sum of digits 2 – 7 + 2 – 8 = -11. Since -11 is divisible by 11, so is 2728.Similarly, for 31415, the alternating sum of digits is 3 – 1 + 4 – 1 + 5 = 10. This is not divisible by 11, so neither is 31415. |
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| 596. |
If the HCF of 65 and 117 is expressible in the form of 65m -117 then the value of m |
| Answer» By Euclid\'s division algorithmb = aq + r,0 ≤ r < a [∵ dividend = divisor × quotient + remainder]⇒ 117 = 65 × 1 + 52⇒ 65 = 52 × 1 + 13⇒ 52 = 13 × 4 + 0∴ HCF (65,117) = 13 ...(i) Also, given that, HCF (65,117) = 65m - 117 ...(ii)From Eqs.(i) and (ii),65m - 117 = 13⇒ 65m = 130⇒ m = 2 | |
| 597. |
If sin A = ¾, calculate cos A and tan A |
| Answer» S t e p - b y - s t e p e x p l a n a t i o n:It is G iv e n\xa0i )\xa0=/* From (1) */====---(1)ii)\xa0=\xa0After cancellation, we get=\xa0---(2)Therefore,\xa0OR | |
| 598. |
Prove Euclid\'s divisions lemma |
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Answer» It\'s already deleted from our syllabus.... Dear it\' s not in syllabus Ex1.1 |
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| 599. |
A line segment is divided in -2:1. Is this possible? |
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Answer» No it is not possible because ratio can not be negative Yes |
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| 600. |
Answers of maths NCERT chapter 5 |
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Answer» In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.(ii) The amount of air present in a cylinder when a vacuum pump removes one fourth\xa0of the air remaining in the cylinder at a time.(iii) The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter.(iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound Interest at 8% per annum.Ans. (i)Taxi fare for 1st km = Rs 15, Taxi fare after 2 km = 15 + 8 = Rs 23Taxi fare after 3 km = 23 + 8 = Rs 31Taxi fare after 4 km = 31 + 8 = Rs 39Therefore, the sequence is 15, 23, 31, 39...It is an arithmetic progression because difference between any two consecutive terms is equal which is 8. (23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, ...)(ii)Let amount of air initially present in a cylinder =\xa0VAmount of air left after pumping out air by vacuum pump =\xa0Amount of air left when vacuum pump again pumps out air=\xa0So, the sequence we get is like\xa0Checking for difference between consecutive terms ...Difference between consecutive terms is not equal.Therefore, it is not an arithmetic progression.(iii)\xa0Cost of digging 1 meter of well = Rs 150Cost of digging 2 meters of well = 150 + 50 = Rs 200Cost of digging 3 meters of well = 200 + 50 = Rs 250Therefore, we get a sequence of the form 150, 200, 250...It is an arithmetic progression because difference between any two consecutive terms is equal. (200 – 150 = 250 – 200 = 50...)Here, difference between any two consecutive terms which is also called common difference is equal to 50.(iv)Amount in bank after Ist year =\xa0… (1)Amount in bank after two years =\xa0… (2)Amount in bank after three years =\xa0… (3)Amount in bank after four years =\xa0… (4)It is not an arithmetic progression because (2) − (1) ≠ (3) − (2)(Difference between consecutive terms is not equal)Therefore, it is not an Arithmetic Progression.2. Write first four terms of the AP, when the first term a and common difference d are given as follows:(i)a = 10, d = 10(ii) a = -2, d = 0(iii) a = 4, d = -3(iv) a = -1, d =\xa0(v) a = -1.25, d = -0.25Ans. (i)\xa0First term = a = 10, d = 10Second term = a + d = 10 + 10 = 20Third term = second term + d = 20 + 10 = 30Fourth term = third term + d = 30 + 10 = 40Therefore, first four terms are: 10, 20, 30, 40(ii)\xa0First term = a = –2 , d = 0Second term = a + d = –2 + 0 = –2Third term = second term + d = –2 + 0 = –2Fourth term = third term + d = –2 + 0 = –2Therefore, first four terms are: –2, –2, –2, –2(iii)\xa0First term = a = 4, d =–3Second term = a + d = 4 – 3 = 1Third term = second term + d = 1 – 3 = –2Fourth term = third term + d = –2 – 3 = –5Therefore, first four terms are: 4, 1, –2, –5(iv)\xa0First term = a = –1, d =\xa0Second term = a + d = –1 +\xa0\xa0= −Third term = second term + d = −+\xa0= 0Fourth term = third term + d = 0 +\xa0=Therefore, first four terms are: –1, −, 0,(v)\xa0First term = a = –1.25, d = –0.25Second term = a + d = –1.25 – 0.25 = –1.50Third term = second term + d = –1.50 – 0.25 = –1.75Fourth term = third term + d= –1.75 – 0.25 = –2.00Therefore, first four terms are: –1.25, –1.50, –1.75, –2.003. For the following APs, write the first term and the common difference.(i) 3, 1, –1, –3 …(ii) –5, –1, 3, 7...(iii)\xa0(iv) 0.6, 1.7, 2.8, 3.9 ...Ans. (i)\xa03, 1, –1, –3...First term = a = 3,Common difference (d) = Second term – first term = Third term – second term and so onTherefore, Common difference (d) = 1 – 3 = –2(ii)\xa0–5, –1, 3, 7...First term = a = –5Common difference (d) = Second term – First term= Third term – Second term and so onTherefore, Common difference (d) = –1 – (–5) = –1 + 5 = 4(iii)First term = a =\xa0Common difference (d) = Second term – First term= Third term – Second term and so onTherefore, Common difference (d) =\xa0(iv)\xa00.6, 1.7, 2.8, 3.9...First term = a = 0.6Common difference (d) = Second term – First term= Third term – Second term and so onTherefore, Common difference (d) = 1.7 − 0.6 = 1.14. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.(i) 2, 4, 8, 16...(ii) 2,\xa0, 3,\xa0...(iii) −1.2, −3.2, −5.2, −7.2...(iv) −10, −6, −2, 2...(v)\xa0(vi) 0.2, 0.22, 0.222, 0.2222...(vii) 0, −4, −8, −12...(viii)\xa0(ix) 1, 3, 9, 27...(x)\xa0a, 2a, 3a, 4a...(xi)\xa0(xii)\xa0(xiii)\xa0(xiv)\xa0(xv)\xa0Ans. (i)\xa02, 4, 8, 16...It is not an AP because difference between consecutive terms is not equal.As4 – 2 ≠ 8 − 4(ii)2,\xa0, 3,\xa0...It is an AP because difference between consecutive terms is equal.Common difference (d) =\xa0Fifth term =\xa0\xa0Sixth term = 4 + ½ =\xa0Seventh term =\xa0Therefore, next three terms are 4,\xa0and 5.(iii)−1.2, −3.2, −5.2, −7.2...It is an AP because difference between consecutive terms is equal.\xa0−3.2 − (−1.2)= −5.2 − (−3.2)= −7.2 − (−5.2) = −2Common difference (d) = −2Fifth term = −7.2 – 2 = −9.2Sixth term = −9.2 – 2 = −11.2Seventh term = −11.2 – 2 = −13.2Therefore, next three terms are −9.2, −11.2 and −13.2(iv)\xa0−10, −6, −2, 2...It isan AP because difference between consecutive terms is equal.\xa0−6 − (−10) = −2 − (−6)= 2 − (−2) = 4Common difference (d) = 4Fifth term = 2 + 4 = 6 Sixth term = 6 + 4 = 10Seventh term = 10 + 4 = 14Therefore, next three terms are 6, 10 and 14(v)\xa0It is an AP because difference between consecutive terms is equal.\xa0Common difference (d) =\xa0Fifth term =\xa0Sixth term =\xa0Seventh term =\xa0Therefore, next three terms are\xa0(vi)\xa00.2, 0.22, 0.222, 0.2222...It is not an AP because difference between consecutive terms is not equal.0.22 − 0.2 ≠ 0.222 − 0.22(vii)\xa00, −4, −8, −12...It is an AP because difference between consecutive terms is equal.\xa0−4 – 0 = −8 − (−4)= −12 − (−8) = −4Common difference (d) = −4Fifth term = −12 – 4 =−16 Sixth term = −16 – 4 = −20Seventh term = −20 – 4 = −24Therefore, next three terms are −16, −20 and −24(viii)\xa0It is an AP because difference between consecutive terms is equal.\xa0Common difference (d) = 0Fifth term =\xa0Sixth term =\xa0Seventh term =\xa0Therefore, next three terms are\xa0(ix)\xa01, 3, 9, 27...It is not an AP because difference between consecutive terms is not equal.3 – 1 ≠ 9 − 3(x)\xa0a, 2a, 3a, 4a...It is an AP because difference between consecutive terms is equal.\xa02a\xa0–\xa0a\xa0= 3a\xa0− 2a\xa0= 4a\xa0− 3a\xa0=\xa0aCommon difference (d) = aFifth term = 4a\xa0+\xa0a\xa0= 5a\xa0Sixth term = 5a\xa0+\xa0a\xa0= 6aSeventh term = 6a\xa0+\xa0a\xa0= 7aTherefore, next three terms are 5a, 6a\xa0and 7a(xi)\xa0a,\xa0a2,\xa0a3,\xa0a4...It is not an AP because difference between consecutive terms is not equal.a2\xa0–\xa0a\xa0≠\xa0a3\xa0−\xa0a2(xii)\xa0\xa0It is an AP because difference between consecutive terms is equal.\xa0Common difference (d) =\xa0Fifth term =\xa0\xa0Sixth term =\xa0Seventh term =\xa0Therefore, next three terms are\xa0(xiii)\xa0It is not an AP because difference between consecutive terms is not equal.(xiv)\xa0It is not an AP because difference between consecutive terms is not equal.(xv)\xa0\xa01, 25, 49, 73...It is an AP because difference between consecutive terms is equal.\xa0=\xa0= 24Common difference (d) = 24Fifth term = 73 + 24 = 97 Sixth term = 97 + 24 = 121Seventh term = 121 + 24 = 145Therefore, next three terms are 97, 121 and 145Exercise 1 Which exercise? |
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