Question

the ratio of the sums of m and n terms of an A.P is m square : n square .show that the ratio of m th and n th terms is ( 2m- 1) :(2n-1)
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Answer 1

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Step-by-step explanation:We know that, sum of x terms of an AP is given by, Sx=x/2(2a+(x-1)d)

Now coming to the question,

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

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

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

(2a+md-d)/(2a+nd-d)=m/n

2an+mnd-nd=2am+mnd-md

2an-nd=2am-md

2an-2am=nd-md

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

2a=d - - - - - - -eq (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,am:an=2m-1:2n-1

Answer 2

Answer:

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