Question

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Answer 1

Question:

The ratio of the sums of m and n terms of an AP is ${m}^{2}:{n}^{2}$. Show that the ratio of the ${m}^{th}$ and the ${n}{th}$ terms is (2m -1):(2n -1).

Answer:

$\large\boxed{\sf{(2m-1):(2n-1)}}$

Step-by-step explanation:

For an AP, we know that,

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

Sum of n terms = n/2 [2a + (n -1)d]

Therefore, we will get,

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

=> [2a + md - d] / [2a + nd - d] = m/n

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

=> 2an - 2am = nd - md

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

=> 2a = d

Now, we have,

Ratio of m th term to nth term,

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

Substituting the value of d, we get,

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

= a [1 + 2m - 2] / a[1 + 2n -2]

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

Hence, the ratio of mth term and the nth term of the A.P is (2m - 1):(2n -1).

Thus, Proved.