Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Pranjal Jangid 5 years, 1 month ago
- 1 answers
Posted by Pranjal Jangid 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
Sum of first 5 multiples of 3:
Sum of an AP is: n/2 × (2a+(n−1)d)
a=3,d=3,n=5
Then,
Sum = 5/2 ×(2×3+(5−1)×(3))
= 5/2 ×(6+12)
= 5/2 ×18
= 5 ×9
= 45
Posted by Pranjal Jangid 5 years, 1 month ago
- 1 answers
Gaurav Seth 5 years ago
let a = first term and d=common difference.
According to question,
7a7 = 11 a11
=>7(a+6d) = 11(a+10d)
=> 7a-11a = 110d-42d
=> -4a = 68d
=>a = -17d ......(1)
Now , a18 = a +17d
but a = -17 d ..by (1)
so put this value , we get,
a18=-17d +17d =0
Hence proved
Posted by Pranjal Jangid 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
a = 1 d = 4-1 = 3
an = a + (n-1)d
a10 = 1 + (10-1)3
a10 = 1 + (9)3
a10= 1 + 27
a10= 28
Posted by Pranjal Jangid 5 years, 1 month ago
- 1 answers
Gaurav Seth 5 years ago
When a, b, and c are real numbers, a ≠ 0 and the discriminant is zero, then the roots α and β of the quadratic equation ax2+ bx + c = 0 are real and equal.
Posted by Pranjal Jangid 5 years, 1 month ago
- 2 answers
Yogita Ingle 5 years, 1 month ago
A quadratic polynomial has maximum three zeroes.
False, A quadratic polynomial has two zeroes.
Posted by Pranjal Jangid 5 years, 1 month ago
- 0 answers
Posted by Pranjal Jangid 5 years, 1 month ago
- 0 answers
Posted by Pranjal Jangid 5 years, 1 month ago
- 2 answers
Yogita Ingle 5 years, 1 month ago
If b2 - 4ac = 0 then the roots will be a real number because b and a are real.
Posted by Pranjal Jangid 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
(cosecA-cotA) (cosecA+cotA)
= cosec2 A - Cot2A
= cot2A + 1 - Cot2A (cot2A + 1 = cosec2A)
= 1
Posted by Pranjal Jangid 5 years, 1 month ago
- 0 answers
Posted by Pranjal Jangid 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
If sino = 3/5
Then, opposite side of the right angled triangle containing angle o = 3 and hypotenuse =5
Using pythagoras theorem, we get the adjacent side = 4
Now we have, coso =Adjacent/Hypotenuse=4/5
And cot o =Adjacent/Opposite= 3/4
3 cot o =3( 3/4) = 9/4
Posted by Pranjal Jangid 5 years, 1 month ago
- 0 answers
Posted by Bharat Kumar 5 years, 1 month ago
- 5 answers
Posted by Khushi Singh Chauhan 5 years, 1 month ago
- 0 answers
Posted by Saidaiah Aludasu 5 years, 1 month ago
- 1 answers
Posted by Jai Sharma 5 years, 1 month ago
- 1 answers
Posted by Shaikh Samim 5 years, 1 month ago
- 1 answers
Gaurav Seth 5 years, 1 month ago
a = - 1.25, d = - 0.25
Let the series be a1, a2, a3, a4 …
a1 = a = - 1.25
a2 = a1 + d = - 1.25 - 0.25 = - 1.50
a3 = a2 + d = - 1.50 - 0.25 = - 1.75
a4 = a3 + d = - 1.75 - 0.25 = - 2.00
Clearly, the series will be 1.25, - 1.50, - 1.75, - 2.00 ……..
First four terms of this A.P. will be - 1.25, - 1.50, - 1.75 and - 2.00.
Posted by Kamaljeet Kaur 5 years, 1 month ago
- 1 answers
Posted by Agatha Kandulna 5 years, 1 month ago
- 2 answers
Posted by Nancy Shetty 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
△ABC ~ △DEF
a(△ABC)/a( △DEF) = AB2 /DE2
64/144 =42/DE2
8/12 = 4/DE
8DE = 48
DE = 48/8 = 6 unit
Posted by Nancy Shetty 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
tan (A+10) = 1
we know that tan 45 = 1
So, tan (A+ 10) = tan45
A + 10 = 45
A = 45 - 10
A = 35
Posted by ?Isha .B 5 years, 1 month ago
- 1 answers
Gaurav Seth 5 years, 1 month ago
Q: Prove that in a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
A n s w e r :
Given: A triangle ABC in which AC2 = AB2 + BC2.
To Prove: ∠B = 90°.
Construction: We construct a ΔPQR right-angled at Q such that
PQ = AB and QR = BC
Proof: Now, from ΔPQR, we have,
Now, in ΔABC and ΔPQR,
<input name="a38150_dofollow" title="Ask a new question relating to this answer" type="submit" value="ask related question" />
Posted by Vaishnavi Ajith 5 years, 1 month ago
- 2 answers
Yogita Ingle 5 years, 1 month ago
Given, a=5,d=3,an=50
⇒a+(n−1)d=50
⇒5+(n−1)3=50
⇒5+3n−3=50⇒3n=48⇒n=16
∴S16=16/2[2a+(16−1)d]=8[2×5+15×3]=440
Hence, n=16,S16=440
Kiran Kaintura 5 years, 1 month ago
Posted by Rishav Negi 5 years, 1 month ago
- 1 answers
Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago
Posted by Gopinath Gopinath 5 years, 1 month ago
- 2 answers
........ ...... 5 years, 1 month ago
Aditya Khohwal 5 years, 1 month ago
Posted by Anurag Pandey 5 years, 1 month ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Gaurav Seth 5 years ago
Given AP is 1,4,7,10,.....
In given AP first term a=1, common difference d=3
Since, Sn= n/2[2a+(n−1)d] ....... sum of n terms
Put n=20 for sum of first 20 terms
S20=20/2 [2(1)+(20−1)3]
S20=10[2+57]
S20=590
0Thank You