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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.
| 43451. |
Ch 5 , Arithmetic progression , Q If 8n =4n - n2, find an |
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Answer» Sum of the first n term = 4n - n^2 Sn = 4n-n^2 then n=1 ,4×1-1^2 =3 the sum of first term is 3 then n=2 , 4×2-2^2=4 the sum of 2 term is 4 then second term = sum of 2 term - sum of 1 term = 4 -3 = 1 then a = 3 , a+d=1 then d =-2 then n term is a+ (n-1)×d = 3+(n-1)×-2 = 5 -2n We have given thatSum of the first n terms = 4n – n2⇒ Sn\xa0= 4n – n2Sn –1\xa0= 4(n – 1) – (n – 1)2= (4n – 4) – (n2\xa0+ 1 – 2n)= 4n – 4 – n2\xa0– 1 + 2n= 6n – n2\xa0– 5∴ nth term = Sn\xa0– Sn – 1= (4n – n2) – (6n – n2\xa0– 5)= 4n – n2\xa0– 6n + n2\xa0+ 5= 5 –2n. |
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| 43452. |
Chapter 14 exercises 14.1 7question solution using direct method |
| Answer» 7. To find out the concentration of SO2\xa0in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:\tConcentration of SO2\xa0( in ppm)Frequency0.00 – 0.0440.04 – 0.0890.08 – 0.1290.12 – 0.1620.16 – 0.2040.20 – 0.242\tFind the mean concentration of SO2\xa0in the air.Solution:To find out the mean, first find the midpoint of the given frequencies as follows:\tConcentration of SO2\xa0(in ppm)Frequency (fi)Mid-point (xi)fixi0.00-0.0440.020.080.04-0.0890.060.540.08-0.1290.100.900.12-0.1620.140.280.16-0.2040.180.720.20-0.2420.200.40TotalSum fi\xa0= 30\xa0Sum (fixi) = 2.96\tThe formula to find out the mean isMean = x̄ = ∑fixi\xa0/∑fi= 2.96/30= 0.099 ppmTherefore, the mean concentration of SO2\xa0in air is 0.099 ppm. | |
| 43453. |
Please answer ex -6.3 question no.10 page no.140... |
| Answer» Class 10 Maths Exercise 6.310. CD and GH are respectively the bisectors of\xa0ACB and\xa0EGF such that D and H lie on sides AB and FE at\xa0ABC and\xa0EFG respectively. If\xa0ABC\xa0FEG, show that:(i)\xa0(ii)\xa0DCB\xa0HE(iii)\xa0DCA\xa0HGFAns.\xa0We have,\xa0ABC\xa0\xa0FEG\xa0A =\xa0F………(1)And\xa0C =\xa0G\xa0C =\xa0G\xa01 =\xa03 and\xa02 =\xa04 ……….(2)[CD and GH are bisectors of\xa0C and\xa0Grespectively]\xa0In\xa0s DCA and HGF, we haveA =\xa0F[From eq.(1)]2 =\xa04[From eq.(2)]\xa0By AA-criterion of similarity, we haveDCA\xa0\xa0HGFWhich proves the (iii) partWe have,DCA\xa0\xa0HGF\xa0Which proves the (i) partIn\xa0s DCA and HGF, we have1 =\xa03[From eq.(2)]B =\xa0E[\xa0DCB\xa0\xa0HE]Which proves the (ii) part | |
| 43454. |
Solve, exercise 5.1,Q.4,(xii) |
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Answer» Thanks (xii) √2, √8, √18, √32\xa0…Here,a2\xa0–\xa0a1\xa0= √8-√2 = 2√2-√2\xa0= √2a3\xa0–\xa0a2\xa0= √18-√8\xa0= 3√2-2√2\xa0= √2a4\xa0–\xa0a3\xa0= 4√2-3√2\xa0= √2Since,\xa0an+1\xa0–\xa0an\xa0or the common difference is same every time.Therefore,\xa0d\xa0=\xa0√2\xa0and the given series forms a A.P.Hence, next three terms are;a5\xa0= √32+√2\xa0= 4√2+√2\xa0= 5√2\xa0= √50a6\xa0= 5√2+√2\xa0= 6√2\xa0= √72a7\xa0= 6√2+√2\xa0= 7√2\xa0= √98 |
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| 43455. |
How many numbers from 100 to 350 which are not divsible by 6 |
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Answer» 50 is wrong Answer is 42 50 According to my thinking answer is \'50\'. |
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| 43456. |
Prove that 4_5root 2 is irrrational number |
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Answer» Who helped lencho Let us assume that 4-5√2 is be rational.=> 4-5√2 = a/b [ where, a,b belongs to Z]=> -5√2 = a/b -4=> √2 = 4/5 -a/bL.H.S = irrational number as √2 is an irrational number .R.H.S = rational numberHence, R.H.S ≠ L.H.STherefore, our assumption is wrong & 4-5√2 is an irrational number. Answer my question warna i delete this app |
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| 43457. |
in isosceles triangle ABC, if AC=BC and AB2=2AC2 then the measure angC ??? |
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Answer» C= 90° Much bmdbmn Triangle ABC is Isoscelesand it is givenAC = BCAB² = 2 AC²= AC² + AC²AC = BC so we can write,AB² = BC² + AC²This is similar to Pythagoras TheoremAB = hypotenuseBC = Perpendicular or BaseAC = Base or PerpendicularThereforeC = 90° |
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| 43458. |
A real number \'a\' is called a zero of polynomial p(x) if: |
| Answer» The constant plynomial f(x) = 0 is\xa0called zero polynomial.A real number \'a\' is called a zero of polynomial p(x) = 0 | |
| 43459. |
Find the two numbers whose sum is 27 and product is182? |
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Answer» 13 &14 13 and 14 13 and 14 13 and 14 Let the first number be\xa0x\xa0and the second number is 27 -\xa0x.Therefore, their product =\xa0x\xa0(27 -\xa0x)It is given that the product of these numbers is 182.Therefore,\xa0x(27 -\xa0x) = 182⇒\xa0x2\xa0– 27x\xa0+ 182 = 0⇒\xa0x2\xa0– 13x\xa0- 14x\xa0+ 182 = 0⇒\xa0x(x\xa0- 13) -14(x\xa0- 13) = 0⇒ (x\xa0- 13)(x\xa0-14) = 0Either\xa0x\xa0= -13\xa0= 0 or\xa0x\xa0- 14 = 0⇒\xa0x\xa0= 13 or\xa0x\xa0= 14If first number = 13, thenOther number = 27 - 13 = 14If first number = 14, thenOther number = 27 - 14 = 13Therefore, the numbers are 13 and 14. |
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| 43460. |
Find a rational number between √2 and √3. |
| Answer» We know that √2=1.414 and√3 =1.732 So rational no between √2and √3 are 1.432 , 1.563 , 1.576 , 1.711 so on.... | |
| 43461. |
Find the sum of all three digit numbers each of which leaves the remainder 2 when divided by 3 |
| Answer» Three digit number which leaves remainder\xa02\xa0when divided by\xa05\xa0is102,107,112,117......997This form an AP whose first term is\xa0=102,\xa0d=5Let\xa0997\xa0is the n th term of AP i.e.\xa0an\u200b=997an\u200b=a+(n−1)d997=102+(n−1)55n=900n=180Sum of all three digits which leaves remainder\xa02\xa0when divided by\xa05\xa0isSn\u200b=n/2 \u200b[2a+(n−1)d]=180/2 \u200b[2×102(180−1)5]=90[204+179×5]=90[204+895]=90×1099=98910 | |
| 43462. |
Find the middle term of theA.P.6,13,20....,216 |
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Answer» How??? 111 |
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| 43463. |
Find the numerator of fractional representation of 0.81 bar |
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Answer» 9/11Numerator is 9 81/100 81/100 |
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| 43464. |
Relationship between zeroes and coefficients of polynomial |
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| 43465. |
What is √6 |
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Answer» Irrational number It is a irrational number √6=2.4494897428...It can\'t be in the form of p/q Therefore it is a irrational number. it is a irrational number |
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| 43466. |
tanq +cotq=4 then the value of the tan^4q+ cot^4q= |
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Answer» Given Equation is tanA + cotA = 4.On squaring both sides, we get= > (tanA + cotA)^2 = (4)^2= > tan^2A + cot^2A + 2tanAcotA = 16= > tan^2A + cot^2A + 2 * tanA * (1/tanA) = 16= > tan^2A + cot^2A + 2 = 16= > tan^2A + cot^2A = 16 - 2= > tan^2A + cot^2A = 14.On squaring both sides, we get= > (tan^2A + cot^2A)^2 = (14)^2= > tan^4A + cot^4A + 2 * tan^4A * cot^4a = 196= > tan^4A + cot^4A + 2 * tan^4A * (1/tan^4A) = 196= > tan^4A + cot^4A + 2 = 196= > tan^4A + cot^4A = 196 - 2= > tan^4A + cot^4A = 194. Given Equation is tanA + cotA = 4.On squaring both sides, we get= > (tanA + cotA)^2 = (4)^2= > tan^2A + cot^2A + 2tanAcotA = 16= > tan^2A + cot^2A + 2 * tanA * (1/tanA) = 16= > tan^2A + cot^2A + 2 = 16= > tan^2A + cot^2A = 16 - 2= > tan^2A + cot^2A = 14.On squaring both sides, we get= > (tan^2A + cot^2A)^2 = (14)^2= > tan^4A + cot^4A + 2 * tan^4A * cot^4a = 196= > tan^4A + cot^4A + 2 * tan^4A * (1/tan^4A) = 196= > tan^4A + cot^4A + 2 = 196= > tan^4A + cot^4A = 196 - 2= > tan^4A + cot^4A = 194. |
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| 43467. |
Ncert ex-13.2 page no.248 question number 3... |
| Answer» It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends.So, the total height of a gulab jamun = 5 cm.Diameter = 2.8 cmSo, radius = 1.4 cm∴ The height of the cylindrical part = 5 cm–(1.4+1.4) cm=2.2 cmNow, total volume of One Gulab Jamun = Volume of Cylinder + Volume of two hemispheres= πr2h+(4/3)πr3= 4.312π+(10.976/3) π= 25.05 cm3We know that the volume of sugar syrup = 30% of total volumeSo, volume of sugar syrup in 45 gulab jamuns = 45×30%(25.05 cm3)= 45×7.515 = 338.184 cm3 | |
| 43468. |
Solve for x and y x+6/y=6 3x-8/y=5 |
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Answer» Above Answer is wrong ... The given equations are: x + 6/y\xa0= 6 ……..(i)\xa03x - 8/y\xa0= 5 ……..(ii)\xa0Putting 1/y\xa0= v, we get:\xa0x + 6v = 6 …….(iii)\xa03x – 8v = 5 ……(iv)\xa0On multiplying (iii) by 4 and (iv) by 3, we get:\xa04x + 24v = 24 ……..(v)\xa09x – 24v = 15 ……..(vi)\xa0On adding (v) from (vi), we get:\xa013x = 39\xa0⇒ x = 3\xa0On substituting x = 3 in (i), we get:\xa03 + 6/y\xa0= 6\xa0⇒ 6/y\xa0= (6 – 3) = 3\xa0⇒ 3y = 6\xa0⇒ y = 2\xa0Hence, the required solution is x = 3 and y = 2.\xa0 |
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| 43469. |
Solve for x and yx+1/2 +y-1/3=8x-1/ 3 +y +1 /2=9 |
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| 43470. |
-√2 and √2 quadratic equation |
| Answer» P(x)=x²-(Sum of zeroes)x+(product of zeroes).Put these values in the bracket as given by question and you will get the correct quadratic equation. | |
| 43471. |
Find the roots of the following quadratic equations. By factorisationx²-3x-10=0 |
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Answer» Gaurav, you always help everyone. I want to know your class. Can you please.... x2\xa0– 3x\xa0– 10=\xa0x2\xa0- 5x\xa0+ 2x\xa0- 10=\xa0x(x\xa0- 5)\xa0+ 2(x\xa0- 5)= (x\xa0- 5)(x\xa0+\xa02)Roots of this equation are the values for which\xa0(x\xa0- 5)(x\xa0+\xa02) = 0∴\xa0x\xa0- 5 = 0 or\xa0x\xa0+ 2 = 0⇒\xa0x\xa0= 5 or\xa0x\xa0= -2 |
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| 43472. |
If alpha and beta are the zeroes of x^2+7x+12 then find the value of 1/alpha+1beta+2alpha beta |
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| 43473. |
Ex.10.2 |
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| 43474. |
Class tenth maths ex 6.4 question no 3 |
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Answer» To prove: ar(ABC)/ar(DBC) =OA/ ODFirst of all we have to do construction.Draw two perpendiculars AP& DM on BcProof : In ∆ ABC & ∆ DCBArea of ∆ Abc÷Area of∆ Dcb= 1/2 × BC×AP÷1/2 × BC × DMArea of∆ Abc/ Area of∆ Dcb= AP/DM.......(1)Now, In ∆Aop & ∆ DomAngle p = Angle M( due to 90°)angle Aop= angle Dom( V.O.A) ∆ AOP~ ∆ DOM( A.A criteria)Ap/ DM= OP/ OM = AO/DOAP/DM= AO/DO....(2)So from 1 & 2 we can say thatar(ABC)/ar(DBC)= OA/DOI hope that it will help you! It\'s too easy Yes I know |
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| 43475. |
If one zeroes of 2x^2-3x+k is reciprocal to other then find value of k |
| Answer» Let, the zeroes of 2x²-3x+k be A and B. B= 1/A 2x²-3x+k = 0 a=2 , b=-3 , c=kSo,Product of zeroes,A×1/A=c/a => 1 = k/2 => k = 2Thus, the value of k is 2. | |
| 43476. |
Divide 6x^3 +13x^2+x-2 by 2x+1 find remainder |
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Answer» Remainder=(-3)=3x²+5+2x+1)(6x³+x-2-6x³+3x²+x-2=10x-2 10x (5)=(-3) Remainder = (-3)=> 3x² + 5 2x+1)6x³ +13x²+ x-2 - 6x³ + 3x² + x -2 = 10 x- 2 - 10 x -2 10x (5) = (- 3) |
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| 43477. |
If two zeroes of the polynomial x^4+x^3-15x^2-29x-6 and 2+_√5 find other zeroes |
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Answer» Two zeroes of given equation arex=2+√5 & x=2-√5=0x-2-√5=0 & x-2+√5=0Both are the factors of the given poly.=[(x-2)-(√5)]×[(x-2)+(√5)]= (x-2)^2 - (√5)^2 [ Since (a+b)(a-b)=a^2-b^2]= x^2 - 4x + 4 - 5= x^2 - 4x - 1 = 0 ...(1) Dividing the given poly. By eq. (1), we get, x^4 + x^3 - 15x^2 - 29x -6 ---------------------------------- x^2 -4x - 1=x^2 +5x + 6=0Above equation is another factor of given equationBy solving it,=x^2 + 3x + 2x + 6 = 0=x(x + 3) + 2(x +3) = 0=(x+3) (x+2) = 0x+3=0 & x+2=0x=-3 & x= -2 x = -3,-2 are another\'s zeroes of the given poly. Given equation.x^4 + x^3 - 15x^2 - 29x - 6 |
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| 43478. |
Form a quadratic polynomial p(y) with sum and product of zeroes are 2 and -3/ 5 |
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Answer» Quadratic polynomial = 5x² - 10x - 3 There are two method a+b =2=c/aab= (-3)/5= (-b)/aWhere a= 1, b= 2, c= (-3/5)Other use this formala to find equationK[x²+ (sum of zeroes )x + (product of zeroes |
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| 43479. |
Find the zeroes of the polynomial 100x^2-81 |
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Answer» 100x²-81 = 0 => 100x² = 81 => x² = 81/100 => x = +/- _/81/100 => x= +/- 9/10So, the zeroes are 9/10 , -9/10 It can be solve by factorzation method |
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| 43480. |
Express 3825 as a product of prime factors |
| Answer» 3825=3×3×5×5×17 | |
| 43481. |
Curved surface area of cubiod |
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Answer» 2h(l+b) Lateral surface area and curved surface area both are same thing Cuboid doesn\'t have curved surface area it has lateral surface area The Curved surface area of cuboid is the sum of the areas of its six faces. The surface area is the total area of all the faces of a 3D shape.\xa0The curved\xa0surface area of a cuboid is the sum of 4 planes of a rectangle, leaving the top (upper) and the base (lower) a. Mathematically, the Curved Surface Area of a cuboid (CSA) is given as:Curved Surface Area of a cuboid (LSA) = 2 (lh + wh) = 2 h (l + w) square units |
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| 43482. |
If p, q are two prime no. then LCM (p, q) is |
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Answer» PQ pq Pq LCM (p,q)= p × q Prime number : A number is called a prime number, if it has no factor other than one and the number itself.LCM : LCM of two or more numbers = product of the greatest power of each common prime factor involved in the numbers, with highest power.SOLUTION : Given : p & q are two Prime numbers.We know that , prime numbers have only two factors 1 and the number itself.Let p = 3 and q = 5The factors of p and q are as follows : p = 3 × 1q = 5 × 1LCM (3,5) = 3 × 5 = 15Hence, LCM(p,q) is pq |
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| 43483. |
If p, q are two co-prime no. then HCF (p, q) is |
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Answer» PQ=1 HCF(p,q)=1 |
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| 43484. |
If d= LCM (36, 198) then the value of d is |
| Answer» A n s w e r is as follows :36 = 2 x 2 x 3 x 3198 = 2 x 3 x 6 x 18 x 11LCM ( 36,198 ) = 2 x 2 x 3 x 3 x 6 x 18 x 11 = 396the value of d = 396 | |
| 43485. |
What is the difference between sequence progression and series |
| Answer» Sequences:\xa0A finite sequence is a sequence that contains the last term such as a1, a2, a3, a4, a5, a6……an.\xa0On the other hand, an infinite sequence is never-ending i.e. a1, a2, a3, a4, a5, a6……an…..Series:\xa0In a finite series, a finite number of terms are written like a1\xa0+ a2\xa0+ a3\xa0+ a4\xa0+ a5\xa0+ a6 +\xa0……an. In case of an infinite series, the number of elements are not finite i.e. a1\xa0+ a2\xa0+ a3\xa0+ a4\xa0+ a5\xa0+ a6 +\xa0……an\xa0+….. | |
| 43486. |
What is the value of cos90\' |
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Answer» 0 -0.448073616 O O -0.44807361612 |
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| 43487. |
How to find out the h. F. C |
| Answer» Smallest factors of a numberHCF=a× b/LCM | |
| 43488. |
zeros of x^2+99x+127 |
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Answer» ohh google Let given quadratic polynomial be p(x) =x2\xa0+ 99x + 127.On comparing p(x) with ax2\xa0+ bx + c, we geta = 1, b = 99 and c = 127 Hence, both zeroes of the given quadratic polynomial p(x) are negative. |
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| 43489. |
In triangle abc right angled at c find cos A, tan A, and cosec B if sinA=24/25 |
| Answer» CosA= 7/25TanA= 24/7CosecB = 25/7 | |
| 43490. |
When was the Nelson Mandela\'s chosen as the president of the Democratic Africa |
| Answer» The African National Congress won a 63% share of the vote at the election, and Mandela, as leader of the ANC, was inaugurated on\xa010 May 1994\xa0as the country\'s first Black President, with the National Party\'s F.W. de Klerk as his first deputy and Thabo Mbeki as the second in the Government of National Unity. | |
| 43491. |
What is nephron? |
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Answer» Nephrons are the microscopic structural and functional unit of the kidney. Nephrons are composed of a renal corpuscle and a renal tubule. The renal corpuscle consists of a tuft of capillaries called a glomerulus and an encompassing Bowman’s capsule.Structure of a Nephrons:\tNephrons are the basic filtering units of kidneys.\tEach kidney possesses large number of nephrons, approximately 1-1.5 million.\tThe main components of the nephron are glomerulus, Bowman’s capsule, and a long renal tubule. Structural unit of kidney |
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| 43492. |
What is Mathematicians associated of number system and area of intersection |
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| 43493. |
29 343 |
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| 43494. |
The nth term of an ap is given-4n+15 . Find the sum of its 20 term |
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Answer» Thanks to all of you On putting n = 1, 2,3 ,…. in eq (1),For n = 1(an) = - 4n + 151 = - 4(1) + 15 a1 = 11\xa0For n = 2a2 = - 4(2) + 15 a2 = - 8 + 15 a2 = 7\xa0Common Difference , d = a2 - a1 d = 7-11 d = - 4\xa0By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]S20 = (20/2)[2 × 11 + (20 -1)(-4)]S20 = 10 [22 + 19 × - 4[S20 = 10 [22 - 76]S20 = 10 × - 54S20 =\xa0- 540 Hence, the sum of first 20 terms of AP is - 540. nth term of an A.P,\xa0(an) = - 4n + 15……..(1)On putting n = 1, 2,3 ,…. in eq (1),For n = 1(an) = - 4n + 151 = - 4(1) + 15 a1 = 11For n = 2a2 = - 4(2) + 15 a2 = - 8 + 15 a2 = 7Common Difference , d = a2 - a1 d = 7-11 d = - 4By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]S20 = (20/2)[2 × 11 + (20 -1)(-4)]S20 = 10 [22 + 19 × - 4[S20 = 10 [22 - 76]S20 = 10 × - 54S20 =\xa0- 540 Hence, the sum of first 20 terms of AP is - 540.\xa0 Kis book ka h Given : nth term of an A.P,\xa0(an) = - 4n + 15……..(1)\xa0On putting n = 1, 2,3 ,…. in eq (1),For n = 1(an) = - 4n + 151 = - 4(1) + 15 a1 = 11\xa0For n = 2a2 = - 4(2) + 15 a2 = - 8 + 15 a2 = 7\xa0Common Difference , d = a2 - a1 d = 7-11 d = - 4\xa0By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]S20 = (20/2)[2 × 11 + (20 -1)(-4)]S20 = 10 [22 + 19 × - 4[S20 = 10 [22 - 76]S20 = 10 × - 54S20 =\xa0- 540 Hence, the sum of first 20 terms of AP is - 540. |
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| 43495. |
If ratio of n terms of two AP is (14n-6):(8n+23). Find the ratio of sum of 17th terms. |
| Answer» 24:19 | |
| 43496. |
In triangles chapter can anyone explain full SSS congruency?? |
| Answer» If three 3⃣ sides of a triangle are equal to the three corresponding sides of other triangle, then triangle??? are congruent. | |
| 43497. |
(X+5)(X+6)=25/(24)^2=0 X=? |
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| 43498. |
1+1-1+....... =9 |
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Answer» 8 7 7 8 8 |
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| 43499. |
If"a" and 1/a are the zeroes of the polynomial the f(x)=4x2-27x+3k2 find the value of k. |
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Answer» 1/3 1/3 |
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| 43500. |
Which term of the A.P. 121,117,113....is its first negative term |
| Answer» A.P : 121, 117, 113...a = 121d = 117- 121=\xa0-4\xa0Let an= 0an\xa0= a + (n-1)d⇒\xa00 = 121 + (n-1)(-4)⇒\xa00 = 121 - 4n + 4⇒4n = 125⇒\xa0n = 125/4⇒\xa0n = 31.25\xa0n must be natural number therefore n= 32\xa0Hence,\xa032nd\xa0term is first negative number. | |