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).
Answers
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:
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 !!