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Posted by Rajbir Yadav 39 minutes ago

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ANSWER

Posted by Muskan Singh an hour ago

- 2 answers

Harshita Barnwal 2 minutes ago

a7 = a + 6d = 49-------------(1)
a17 = a + 16d = 289------------(2)
Subtract (1) from (2) ;
a + 6d = 49
a + 16d = 289
(-) (-) (-)
—————————
-10d. = -240
10d = 240
: [ d = 24 ]
Putting the value of d in (1) ;
a + 6d = 49
a + 6 × 24 = 49
a + 144 = 49
a = 49 - 144
[ a = -95 ]
An is not given in your question otherwise the formula of an is "a+(n-1)d"............

Posted by Khushi Keshri an hour ago

- 14 answers

Anshu Kumari an hour ago

I m also new here.....
....and i also want to make new frnd here

Posted by Su Ghosh an hour ago

- 6 answers

Kamlesh Dubey an hour ago

Yes u can take physics in class 11th but you cannot take maths as an extra subject also if you want so you have to appear in standard maths

Posted by Niraj Mali 2 hours ago

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Posted by Lakhan Bairagi 3 hours ago

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Posted by Zaid Ali 4 hours ago

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Posted by Rajeev Mishra 4 hours ago

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Posted by Muskan Sharma 4 hours ago

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Posted by Dhriti Pandey 5 hours ago

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Posted by Sanket Sundar 6 hours ago

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Posted by Sandeep Singh Sankhla 9 hours ago

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Posted by Roshikumar Tongbram 9 hours ago

- 2 answers

Khushi Simaiya 9 hours ago

X,2x ,6x are the angles..as assumed.
Now x+2x+6x =180
Therefore 9x=180
x=20
Now substituting the valye of x in assumption we get all three angles they are 20,40,120..
Hope it helps u mate..

Amit Kumar 9 hours ago

Let the angles be x,2x and 6x.
Then, x+2x+6x=180
9x=180
x=20
Therefore; angles are 20, 20×2=40, 20×6=120.

Posted by Raj Chaudhari 10 hours ago

- 1 answers

Akanksha Tiwari 9 hours ago

If we Choose basic Math we will not able to take any stream related to math. But if we choose standard math we will able to take stream related to math.

Posted by Tharki Baba Om!! 12 hours ago

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Posted by Amey Gupta 13 hours ago

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Ruchi Kanojiya 12 hours ago

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Posted by Sristi Pandey 13 hours ago

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Gaurav Seth 13 hours ago

**Answer:**

The value of k is 3.

**Step-by-step explanation:**

The given quadratic equation is

It is given that one root of this quadratic equation is reciprocal of the other.

Let the roots be .

We know that,

....(1)

....(2)

Using equation (2) we get

Multiply both sides by 3.

**Therefore the value of k is 3.**

Posted by Amrit Keshari Patra 16 hours ago

- 2 answers

Vanshika Goyal 10 hours ago

n=1
5-2*1=3
Similarly when
n=2
5-2*2=1
n=3
5-2*3=-1 and so on...........
This forms an AP =3,1,-1
a=3 & d=-2
Sn =n/2[3+5-2n]
=n/2[8-2n]
=n/2*2(4-n)
=n(4-n)

Posted by Anku Kumar Singh 21 hours ago

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Posted by Siraj Kanjaria 1 day ago

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Posted by Sanket Sundar 11 hours ago

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Sia 🤖 11 hours ago

Let a be the first term and d be the common difference of the given A.P. Then,

S_{m} = S_{n}

{tex}\Rightarrow \quad \frac { m } { 2 } \{ 2 a + ( m - 1 ) d \} = \frac { n } { 2 } \{ 2 a + ( n - 1 ) d \}{/tex}

{tex} \Rightarrow{/tex}_{ }2a (m - n) + {m (m - 1) - n (n - 1)} d = 0

{tex} \Rightarrow{/tex} 2a (m - n) + {m^{2} - m - n^{2} + n}d = 0

{tex} \Rightarrow{/tex}_{ }2a (m - n) + {(m^{2} - n^{2}) - (m - n)} d = 0

{tex} \Rightarrow{/tex}2a (m - n) + {(m^{ } - n) (m +n) - (m - n)} d = 0

{tex} \Rightarrow{/tex}_{ }(m - n) {2a + (m + n - 1)d} = 0

{tex} \Rightarrow{/tex} 2a + (m + n - 1)d = 0_{ }{tex} [ \because m - n \neq 0 ]{/tex} ...(i)

Now, {tex} S _ { m + n } = \frac { m + n } { 2 } \{ 2 a + ( m + n - 1 ) d \} = \frac { m + n } { 2 } \times {/tex}0 = 0 [Using (i)]

Posted by Bir Singh 1 day, 1 hour ago

- 4 answers

Bhumika Jindal 1 day ago

I think this is not any A.P. because their difference is not same

Posted by Bir Singh 1 day, 1 hour ago

- 4 answers

Raj Kumar 1 day, 1 hour ago

(A) yes it is an ap bcoz its common difference is in all terms

Posted by Heart ♥ Hacker 1 day, 1 hour ago

- 5 answers

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