Question

jo aata bata do waise 35 hi batao ​

Answer 1

Step-by-step explanation:

34)

If the first term of an AP is a and the common difference is d and the number of terms is n then the general term or nth term = an =a+(n-1)d

Given that :

pth term of the AP = q

=>a+(p-1)d = q-----(1)

q th term of the AP = p

=>a+(q-1)d = p------(2)

On Subtracting (2) from (1) then

=>a+(p-1)d - a-(q-1)d = q-p

=>(a-a)+[(p-1)-(q-1)]d = (q-p)

=>0+(p-1-q+1)d = (q-p)

=>(p-q)d = (q-p)

=>-(q-p)d = (q-p)

=>d = (q-p)/-(q-p)

=>d = 1/-1

=>d = -1

Common difference is -1

On Substituting the value of d in (1) then

=>a+(p-1)(-1)=q

=>a-p+1=q

=>a = p+q-1

First term of the given AP is p+q-1

Now (p+q)th term of the AP is

=>a+(p+q-1)d

=>p+q-1+(p+q-1)(-1)

=>p+q-1-p-q+1

=>(p-p)+(q-q)+(1-1)

=>0+0+0

=>0

Therefore, (p+q)th term of the AP is zero.

35)

Given quadratic equation is

(1+m^2)x^2+2mcx+c^2-a^2 = 0

On comparing with the standard quadratic equation ax^2+bx+c=0

a = (1+m^2)

b=2mc

c=(c^2-a^2)

Given that

The given equation has equal real roots then

its discriminant must equal to zero

=>b^2-4ac = 0

=>(2mc)^2 -4(1+m^2)(c^2-a^2)=0

=>4m^2c^2-4c^2-4m^2c^2+4a^2+4m^2a^2 = 0

=>4a^2+4m^2a^2-4c^2 =0

=>4(a^2+m^2a^2-c^2)=0

=>a^2+m^2a^2-c^2=0/4

=>a^2+m^2a^2-c^2=0

=>a^2+m^2a^2 = c^2

=>a^2(1+m^2)=c^2

Therefore,c^2=a^2(1+m^2)

Hence, Proved

Used formulae:-

• If the first term of an AP is a and the common difference is d and the number of terms is n then the general term or nth term = an =a+(n-1)d
• The given equation has equal real roots then
• its discriminant must equal to zero
• =>b^2-4ac = 0