Question

solve this fast
❌❌spamming not allow here❌❌​

Answer 1

Ques.11

Answer:

Given:

First term of AP=a

Second term of AP=b

Third term of AP=c

Common difference of AP=b-a

Solution:

Now,

$\sf{a_n=last\:term=a+(n-1)d}$

$\sf{\implies}{c=a+(n-1)(b-a)}$

$\sf{\implies}{c-a+b-a=n}$

___________________________

$\sf{b-a}$

___________________________

$\sf{{\dfrac{b+c-2a}{b-a}}=n}$

Sum of AP=$\sf{S_n}{=}{\dfrac{n}{2}}{[a+a_n]}$

$\sf{S_n=}{\dfrac{(a+c)(b+c-2a)}{2(b-a)}}$

Ques.12

Answer:

Given:

$\sf{S_p=S_q}$

Solution:

$\sf{\implies}{\dfrac{p}{2}}{(2a+(p-1)d)}{=}{\dfrac{q}{2}}{(2a+(q-1)d)}$

$\sf{\implies}{p(2a+(p-1)d)=q(2a+(q-1)d)}$

$\sf{\implies}{2ap+p^2d-pd=2aq+q^2d-qd}$

$\sf{\implies}{2a(p-q)+(p+q)(p-q)d-d(p-q)=0}$

$\sf{\implies}{(p-q)[2a+(p+q)d-d]=0}$

$\sf{\implies}{2a+(p+q)d-d=0}$

$\sf{\implies}{2a+((p+q)-1)d=0}$

Hence,$\sf{S_{p+q}=0}$

Ques.13

Answer:

Let the first term of AP = a

common difference = d

To show:

(m+n)th term is zero or a + (m+n-1)d = 0

Solution:

mth term = a + (m-1)d

nth term = a + (n-1) d

Given that m{a +(m-1)d} = n{a + (n -1)d}

⇒ am + m²d -md = an + n²d - nd

⇒ am - an + m²d - n²d -md + nd = 0

⇒ a(m-n) + (m²-n²)d - (m-n)d = 0

⇒ a(m-n) + {(m-n)(m+n)}d -(m-n)d = 0

⇒ a(m-n) + {(m-n)(m+n) - (m-n)} d = 0

⇒ a(m-n)  + (m-n)(m+n -1) d  = 0

⇒ (m-n){a + (m+n-1)d} = 0  

⇒ a + (m+n -1)d = 0/(m-n)

⇒ a + (m+n -1)d = 0

Answer 2

He bgh..................

If the points (x, 1), (3, y), (-2, 3) and (-3, -2) be the vertices of a parallelogram, then find the value of

2x + y.