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a₁, a₂, a₃ , ......... an are in AP. so, a₂ - a₁ = a₃ - a₂ = a₄ - a₃ = ...... = an₋₁ - an = d also, a₂ = a₁ + d, a₃ = a₂ + d , a₄ = a₃ + d ...... , an = an₋₁ + (n-1)d

Now, LHS = 1/a₁.a₂ + 1/a₂a₃ + 1/a₃a₄ + ..... + 1/an₋₁.an = 1/a₁(a₁ + d) + 1/a₂(a₂ + d) + 1/a₃(a₃ + d) + ....... 1/an₋₁(an₋₁ + d) = 1/d[ 1/a₁ - 1/(a₁ + d) + 1/a₂ - 1/(a₂ + d) + 1/a₃ - 1/(a₃ + d) + ....... +1/an₋₁ - 1/an ]= 1/d [1/a₁ - 1/a₂ + 1/a₂ - 1/a₃ + 1/a₃ - 1/a₄ + ......... + 1/an₋₁ - 1/an ]= 1/d[ 1/a₁ - 1/an ]= 1/d(an - a₁)/a₁.an

We know, an = a₁ + (n-1)d an - a₁ = (n-1)d

= 1/d × (n-1)d/a₁an = (n-1)/a₁an = RHS

Hence proved

sin(A+B) ≡ sin(A)cos(B) + sin(B)cos(A)

⇒ sin(75°) = sin(30+45)° = sin(30°)cos(45°) + sin(45°)cos(30°)

= ½ × 1/√2 + 1/√2 ×(√3)/2 = 1/(2√2) + (√3)/(2√2)

= (1+√3)/(2√2)

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one bundle of stick = 10 sticks. 10+8 = 18 1 ) 10 + 5 = 15 2 ) 10 + 6 = 16

15 and 16 is the answer of the following

15&16 answer........

1)152)16 is the best answer

1)152)16 is the right answer

1)152)16 is the correct answer

correct answer is 1516

here buch of match stick + 8 sticks =18 which means bunch = 18-8=10

que 1 bunch +5 =10+5= 15

que 2bunch +6 =10+6 =16

one bundle of stick = 10 stick 10+8 = 181) 10 + 5 = 152)10 + 6 =16

one bundle of stick 10 stick 10 with 8 = 18 1 . 10+5=152. 10+6=16

15,16 is the answer because the bundle of matchsticks contain 10 matchsticks each so bundle of matchsticks + 5 is equals to 15 and bundle of matchstick plus 6 is equals to 16

15 & 16 is the best answer of the following question

The best answer is

1) 15 2) 16

5+10=15 6+10=16 🛐🔄☣☣🚯🚯🐏⛑⛑😳😳😳⛑🐏😠😧😧😠😤😫🤡🇯🇴😫🤥😚🇮🇶🇰🇼🇰🇵

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seema age 11 and sudesh 25

2880

Profit = 20% = 360 Rs

So, Cost Price = 360 / 20 * 100 = 1800 Rs

And Sales Price = 1800 + 360 = 2160 Rs

Discount = 25% of marked price

So Sales Price = (100 - 25) % = 75% of marked price

=> 2160 = 75% of marked price

=> marked price = 2160 / 75 * 100 = 2880 Rs

tq

Thank you for giving answer..

Manufacturer's gain= 6%

SP of the manufacturer= 106% of Rs.x

=Rs. 53x/50

Wholesale dealer's SP=108% of 53x/50

{53x/50*108/100}= {53x/50*27/25}

Retailer's SP=110% of {53x/50*27/25}

={53x/50*27/25*110/100}

={53x/50*27/25*11/10}

Retail price=Rs. 15741

53x/50*27/25*11/10=15741

x=15741*50*25*10/53*27*11

x=12500

So, the production cost of geyser is Rs. 12500

Let A age be xThen A's grand father age = 63 + xA's father age = 63 + x - 35

As per given conditionx + (63 + x) + (28 + x) = 1123x + 91 = 1123x = 112-91 = 21x = 21/3 = 7

Therefore,Age of A = 7 yearsAge of grand father = 63 + 7= 70Age of father = 28 + 7 = 35

let fathers age be X and daughters age be ynow as y=3x which was 4 years backso present ages will bex+4 and 3x+4 of father and daughter respectivelyin 8 years time2(X+4+8)=3x+4+8hence2x+24=3x+12x=12hence age of daughter is 12and father is 3x+4=3*12+4=40

x+y-12 is the right answer of the question 😊😊

x+y-12 is the correct answer

x+y-12 is the right answer

x + y - 12 the right answer.

x+y+16 its equation can bex+4+y+4=0x+y+16=0

200*S.P.=400*C.P.S.P.=2*C.P.so it is gainand gain percentage=(S.P.-C.P.)/C.P.*100=(2*C.P.-C.P.)/C P.*100=100%

Let in an AP first term = a Common difference = d9th term = 19a + 8d = 19----(1)19th term =3*(6th term)a + 18d = 3(a+5d)a + 18d = 3a+15d0 = 3a+15d-a-18d2a - 3d = 0-----(2)subtract (2) from 2*(1)-19d = -38d = 2substitute d=2 in (1)a=3

Therefore a=3, d=2Required AP is3, 5, 7, 9,....

Answer

Given: Mean of three students = 40 kg

The three students are Ranga, Rahim and Reshma, where

Ranga’s weight = 46 kg

Let Rahim’s weight = x

And Reshma’s weight = y [∵Rahim’s weight = Reshma’s weight, according to the question]

Mean is given by

∵Mean = 40]

⇒120 = 46 + 2x

⇒2x = 120 – 46 = 74

⇒x = 74/2

⇒x = 37

Thus, Rahim’s weight = 37 kg

12,20,28,36,44

Area of Reactangular field = 440 m^2

Length of Reactangular field = 22 mLet breadth of Reactangle field = x

Area = length * breadth440 = 22 * xx = 440/22 = 20 m

Breadth = 20 m

Perimeter = 2(length + breadth)= 2(22 + 20)= 2*42= 84 m

the answer is the question ofanswer is 84 m

Let the no. be nWhen no. Increased by 10%= n + n*(10/100) = n (1 + 1/10)= n*(11/10)

When no. decreased by 10%= n - n*(10/100)= n(1-1/10) = n*(9/10)

Givenn*(11/10) + n*(9/10) = 100n(20/100) = 100n(1/5) = 100n = 500

Number is 500

Resultant = √3²+4²+2*(3)*(4)(cos(30) = √9+16+24*√3/2 = √(25+12√3)

Let the largest portion be as xthen according. to the question ......2:5:7 will be assumed as......2+5+7= 142/14 of x = 40= x = 40*14/2= x =280kgthen respectively112kgand 80kgtherefore.....280 - 40= 220 kg

CSA=2/3×[2(pi×r²)]2pi×r×h=2/3×[2(pi×r²)]3×2pi×r×h=2×[2(pi×r²)]3×2pi×r×h=2×2pi×r²3×r×h=2r²r×3×15=2r²(3×15)/2=r²/r45/2=rr=22.5cm

Hence radius of the cylinder is 22.5cm

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Let the radii be 2x and 3xlet the height be 5y and 3y

Ratio of Lsa = 2πrh/2πRH

= 2x*5y/3x*3y

= 10/9

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Weight is proportional to mass/(radius) ^2

Therefore weight at other planet = 36x(1/9)/(1/2)^2

= 4 x 4= 16.

The perimeter is sum of all sides.So , perimeter= 3+4+2+3=12cm

find the hcf of 18 24 32