Question

solve it fassssssss​

Answer 1

Answer:

Heya user................ ❤

As we know that, sum of x terms of an AP is given by Sx = x/2 × ( 2a + (x-1)d)

Hence,

Sum of m terms of AP = m/2 (2a +( m-1) d)

& Sum of n terms of Ap = n/2 (2a + (n-1) d)

Given that,

Sm:Sn = m^2 : n^2

Therefore, [ m/2 (2a +( m-1) d) ] / [ n/2 (2a+ (n-1) d)] = m^2 :n^2

=> 2an -2am =nd - md

=> 2a (n-m) = d(n-m)

Therefore, 2a = d............(1)

We know that, xth term of an AP is given by, ax=a+(n-1) d

So, the mth term of AP is am=a+(m-1)d

nth terms of AP is an=a+(n-1)d

So, the ratio of mth and nth term of AP is

am/an=[a+(m-1)d] /[2a+(n-1)d]

=(a+(m-1)2a)/(a+(n-1)2a) (from eq1)

=(a+2am-2a)/(a+2an-2a)

=(2am-a)/(2an-a)

=a(2m-1)/a(2n-1)

=(2m-1)/(2n-1)

So,m^th : n^th =2m-1:2n-1

(proved)

Hope it's helpful...... ☺

Answer 2

Answer:

hii !

Ur answer is attached above...