CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Let A =first term of the ap Let d =common difference of the ap Now a=A+(p-1)d............1 b=A+(q-1)d............2 c=A+(r-1)d.............3 Subtracting 2 from 1,3 from 2 and 1 from 3 a-b=(p-q).d..........4 b-c=(q-r).d............5 c-a=(r-p).d............6 Multiplying 4,5,6 by c,a,b c.(a-b)=c.(p-q).d...........4 a.(b-c)=a.(q-r).d...........5 b.(c-a)=b.(r-p).d.............6 a(q-r)d+b(r-p)d+c(p-q)d=0[(a(q-r)+b(r-q)+c(p-q)d=0 Now, since d is common difference it should be non zero Hence a(q-r)+b(r-q)+c(p-q)=0.

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Namrata Nath 7 years, 9 months ago

Page number 341

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Namrata Nath 7 years, 9 months ago

Ye u-like Ka hai

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How many questions are asking from chapter-Triangle in CBSE Board examination?

Posted by Samiksha Srivastava 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Nancy Rajput 7 years, 9 months ago

Check blue print

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The PHQ ethanol term of an ap are a b and c respectively show that a cube minus or plus b a minus b + c b minus Q = 20

Posted by Madhav Gautam 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If the 7times the 7th term of an ap its equal to 11times the 11th term of an ap then find 18th term

Posted by Kunal Arya 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Rashi Malik 7 years, 9 months ago

7th term = (a+6d) 11th term = (a+10d) (a+6d)7 since 7th term of the ap is 7 times and the same for 11th term (a+6d)7=(A+10d)11 (given) 7a + 42d =11a +110d (on solving) 7a - 11a = 110d - 42d -4a = -68d a= -17d we know 18th term = (a+17d)................................. (i) now we substitute the value of a in (i) which emplies, -17d + 17d = 0

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Who is Aditya Kumar? Am I know you.

Posted by Namrata Nath 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

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If the points (p,q) (m,n)and (p-m)(q-n)are collinear ,then prove that pn-qm

Posted by Sidharth Rana 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given points are collinear. Therefore

[p {tex}\times{/tex} n + m(q - n) + (p - m) q] - [m {tex}\times{/tex} q + (p - m) n + p (q - n)] = 0
{tex}\Rightarrow{/tex} (pn + qm - mn + pq - mq) - (mq + pn - mn + pq - pn) = 0
{tex}\Rightarrow{/tex} (pn + p q - mn) - (mq - mn + pq) = 0
{tex}\Rightarrow{/tex} pn - mq = 0
{tex}\Rightarrow{/tex} pn = qm

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Hiiii everyone

Posted by Ziya Sharma 7 years, 9 months ago

CBSE > Class 10 > Mathematics

4 answers

Anjanee Kumari 7 years, 9 months ago

???????

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Anjanee Kumari 7 years, 9 months ago

Chat

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Ziya Sharma 7 years, 9 months ago

Kya kar rhi ho

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Anjanee Kumari 7 years, 9 months ago

Hey!

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Can we easily understand the trigonometry

Posted by Nishu Jaat 7 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Cutiee Princess 7 years, 9 months ago

Ya u can. U just read the formulas n apply it in the given question u will get ur answer

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Namrata Nath 7 years, 9 months ago

If you get the answer tell me also

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1+2+3+4+5.................+100=?

Posted by Namrata Nath 7 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Namrata Nath 7 years, 9 months ago

Thanks buddy

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Cutiee Princess 7 years, 9 months ago

Here a= 1 , d = 1 , l = 100 , Sn = n/2 (a+l ) = 100/2(1+100) = 50 * 101 = 5050

3Thank You

Haardik Ravat 7 years, 9 months ago

a=1 d=1 l=100 Sn = n/2(a+l) =50×101 =5050

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ANSWER

A fraction becomes 9÷11 if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5÷6. Find the fraction

Posted by Khushi Bhatt 7 years, 9 months ago

CBSE > Class 10 > Mathematics

4 answers

Cutiee Princess 7 years, 9 months ago

Ans is 7/9

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Tanu Priya 7 years, 9 months ago

Sorry 7/9 is the correct

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Tanu Priya 7 years, 9 months ago

X= -43 and y=51

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Yuvraj Singh 7 years, 9 months ago

7/9

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What is the ordinate of a point on X - axis ?

Posted by Sakshi Chauhan 7 years, 9 months ago

CBSE > Class 10 > Mathematics

6 answers

Ashutosh Kumar 7 years, 9 months ago

1Thank You

Cutiee Princess 7 years, 9 months ago

( x , 0)

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Yuvraj Singh 7 years, 9 months ago

1Thank You

Sakshi Chauhan 7 years, 9 months ago

Thank u

1Thank You

Prince King 7 years, 9 months ago

0 x -axis

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Nancy Rajput 7 years, 9 months ago

X, 0

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Posted by Ragav Rohi 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

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if a has probality4/3 and b has probality8/5 what that c has neither chances

Posted by Navi Saggu 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Nikhil Raj Mehta 7 years, 9 months ago

Dono probability ko sum karke 1 se subtact kar do... Answer aa jayega..

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ABC ia triangle coordinates of whose vertex a aer (0,-1).d and e respectively are the mid points of the side ab and AC and their coordinates are (1,0) and (0,1) respectively.if f is the midpoint of BC find the area of triangle ABC and triangle def

Posted by Vijay Kumar 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

According to the question, A(0, - 1),D(1, 0) and E(0,1).

Let coordinates of B and C are (x₂, y₂) and (x₃, y₃) respectively.
D is mid-point of AB,
{tex}\therefore{/tex} 1 = {tex}\frac { 0 + x _ { 2 } } { 2 }{/tex}
{tex}\Rightarrow{/tex} x₂ = 2
and 0 = {tex}\frac { - 1 + y _ { 2 } } { 2 }{/tex}
{tex}\Rightarrow{/tex} y₂ = 1
{tex}\therefore{/tex} Coordinates of B are (2, 1)
E is mid-point of AC
{tex}\therefore{/tex} 0 = {tex}\frac { 0 + x _ { 3 } } { 2 }{/tex}
{tex}\Rightarrow{/tex} x₃ = 0
and 1 = {tex}\frac { - 1 + y _ { 3 } } { 2 }{/tex}
{tex}\Rightarrow{/tex} y₃ = 3
{tex}\therefore{/tex} Coordinates of C are (0, 3)
Area of {tex}\triangle{/tex}ABC
{tex}= \frac { 1 } { 2 }{/tex} {tex}[0(1 - 3) + 2(3 + 1) + 0(-1 + -1)]{/tex}
{tex}= \frac { 1 } { 2 }{/tex} {tex}\times{/tex} 8 = 4 sq. units
F is mid-point of BC
{tex}\therefore{/tex} Coordiantes F are {tex}\left( \frac { 2 + 0 } { 2 } , \frac { 1 + 3 } { 2 } \right){/tex}, i.w, (1, 2).
Area {tex}\triangle{/tex}DEF
{tex}= \frac { 1 } { 2 }{/tex} {tex}[1(2 - 1) + (1 - 0) + 0(0 - 2)]{/tex}
{tex}= \frac { 1 } { 2 }{/tex} [1 + 1] = 1 sq. units

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The sum of first 30 terms and 40 terms of an A.P are 2265 and 4020 find common difference

Posted by Kajal Bhatotia 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Rex Cartin 7 years, 9 months ago

d=5

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Kya maths k paper hard aayega bata do

Posted by Viahal Pandat 7 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Tanu Priya 7 years, 9 months ago

Hloo garcy how are y

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Ankit Dey 7 years, 9 months ago

No very sorry

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Shanaya Dabas 7 years, 9 months ago

May be

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If three cards are drawn from a pack of 52 cards ,what is the chance that all will be kings

Posted by Vikash Kumar 7 years, 9 months ago

CBSE > Class 10 > Mathematics

4 answers

Satyam Garg 7 years, 9 months ago

i thought it would be3/52

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Nimesh Chauhan 7 years, 9 months ago

Sorry it's 1/13

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Anmol Jain 7 years, 9 months ago

4/52 = 1/13

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Nimesh Chauhan 7 years, 9 months ago

4/52=1/3

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The angle of depression of the top and bottom of a building 50m high as observed from the top of a tower are 30°and 60°, respectively. Find the high of the tower and also the horizontal distance between the building and the tower.

Posted by Naveen Singh 7 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Garima Gupta 7 years, 9 months ago

Let the height of the tower is h and horizontal distance between building and tower is x. X is equals to under root 3(h-50) H equals to under root 3x. .........1 Put the value of x then we get h is equal to 75 m Now put h in equation 1 then x is equal to 25 multiply under root 3.

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Rex Cartin 7 years, 9 months ago

In ΔBTP, tan 30° = TP/BP 1/√3 = TP/BP BP = TP√3 In ΔGTR, tan 60° = TR/GR √3 = TR/GR GR = TR/√3 As BP = GR TP√3 = TR/√3 3 TP = TP + PR 2 TP = BG TP = 50/2 m = 25 m Now, TR = TP + PR TR = (25 + 50) m Height of tower = TR = 75 m Distance between building and tower = GR = TR/√3 GR = 75/√3 m = 25√3 m I Hope this helps you

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What value of tan thita ?

Posted by Rahul Kumar 7 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Deependra Singh 7 years, 9 months ago

Pappu ho itna bhi nahi pata

1Thank You

Nancy Rajput 7 years, 9 months ago

Opposite of theta/adjacent of theta

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Divide 39 in two parts such that their product is 324. How to solve

Posted by Rinkle Prajapati 7 years, 9 months ago

CBSE > Class 10 > Mathematics

4 answers

Rex Cartin 7 years, 9 months ago

Let first number be x Let the second number be 39-x According to the question, (x) (39-x) =324 39x-x²=324 -x²+39x-324=0 -(x²-39x+324) x²-39x+324=0 Now splitting the middle term x²-27x+12x+324=0 x(x-27)+12(x-27) x=27 or x=-12 . . . . I hope this helps you ?

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Shanaya Dabas 7 years, 9 months ago

I2+27=39 and 12×27=324

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Rinkle Prajapati 7 years, 9 months ago

No its right question

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Mr Lovely 7 years, 9 months ago

I think its wrong question

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The pointsA (4,-6),B (7,2),C,(0'9) ,D(-3,5) form a parallelogram find the length of the altitude of the parallelogram on the base AB

Posted by Ravi Rawat 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

How to solve trigonometry sum

Posted by Rinkle Prajapati 7 years, 9 months ago

CBSE > Class 10 > Mathematics

4 answers

Manik Pandit 7 years, 9 months ago

By using ur own hand ???

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Nishant Verma 7 years, 9 months ago

By pen

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Mayank Sahu 7 years, 9 months ago

By inteligente mind??

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Nancy Rajput 7 years, 9 months ago

by identities

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How to solve math paper quickly

Posted by Rinkle Prajapati 7 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Mayank Sahu 7 years, 9 months ago

By fast speed nd intelligent mind...??

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Shanaya Dabas 7 years, 9 months ago

Start with 4 marks questions

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Divide 39 into two parts such that their product is 324.

Posted by Nancy Rajput 7 years, 9 months ago

CBSE > Class 10 > Mathematics

9 answers

Nancy Rajput 7 years, 9 months ago

fine

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Mr Lovely 7 years, 9 months ago

How r u

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Nancy Rajput 7 years, 9 months ago

hii

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Krishna Pandey 7 years, 9 months ago

27 and 12

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Mr Lovely 7 years, 9 months ago

Hii

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Nancy Rajput 7 years, 9 months ago

thanks

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Cutiee Princess 7 years, 9 months ago

27+12=39 n 27*12= 324

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Shanaya Dabas 7 years, 9 months ago

The number are 27 and 12

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Srijan Kumar 7 years, 9 months ago

27 and 12

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If pth term of an AP is a and qth term of an AP is b then prove that sum of its (p+q)th term is p+q÷2(a+b+(a-b)÷p-q).

Posted by Evlia Saiju 6 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 5 months ago

Let A be the first term and D be the common difference ofthe given AP. Then,
T_p= a {tex}\Rightarrow{/tex} A + (p - 1) D = a ...(i)
and T_q = b {tex}\Rightarrow{/tex} A + (q - 1) D = b ...(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(p - q) D = a - b {tex}\Rightarrow{/tex} D = {tex}\frac { a - b } { p - q }{/tex} ...(iii)
On adding Eqs. (i) and (ii), we get
2A + (p + q - 2) D = a + b
{tex}\Rightarrow{/tex} 2A + pD + qD - 2D = a + b
{tex}\Rightarrow{/tex} 2A + pD + qD - D = a + b + D
{tex}\Rightarrow{/tex} 2A + (p + q - 1) D = a + b + D
2A + (p + q - 1) D = a + b + {tex}\left( \frac { a - b } { p - q } \right){/tex} [from Eq. (iii)] ...(iv)
Now, S_p+q = {tex}\frac { p + q } { 2 }{/tex} [2A + (p + q - 1) D]
= {tex}\frac { p + q } { 2 }{/tex} [a+ b + {tex}\frac { a - b } { p - q }{/tex}] [from Eq. (iv)]
Hence proved.