Question

The ratio of sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of the mth and nth terms is (2m – 1) : (2n – 1).

Answer 1

Answer:

Step-by-step explanation:

ratio of sum = A:P = m2:n2

numbers are m and n

ratio of mth and nth term= (2m-1) : (2n-1)

mth term = (2m-1)x

nth term = (2n-1)x

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

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

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

Ratio of m th term to nth term:

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

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

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

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

Answer 2

Answer:

Step-by-step explanation:

Let a be the first term and d be the common difference of the given AP

A.T.Q

Sum of m terms / Sum of n terms  = m²/n²

⇒m/2[2a + (m-1)d] / n/2[2a +(n-1)d] = m²/n²

⇒2a + (m-1)d / 2a + (n-1)d = m/n

⇒n[2a + (m-1)d] = m[2a + (n-1)d]

⇒2na +nmd -nd = 2ma + nmd -md

⇒2na -nd = 2ma - md

⇒2na -2ma = nd - md

⇒2a(n-m) = (n-m)d

⇒2a =d ------------- (1)

Now , we know that nth term = a + (n-1)d

Therefore, mth term / nth term = a+(m-1)d / a+(n-1)d

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

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

                                                  = 1+2m -2 / 1+2m -2

                  mth term / nth term= 2m-1 / 2n-1

                           HENCE PROVED !!