# The ratio of the sums of m and n terms of an A.P. is m^{2} : n^{2} . Show that the ratio of m^{th} and n^{th} term is (2m – 1) : (2n – 1)

**Solution:**

Let a and d be the first term and the common difference of the given arithmetic progression respectively.

According to the given condition,

⇒ Sum of m terms/Sum of n terms = m^{2}/n^{2}

⇒ 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 ....(1)

Putting m = 2m - 1 and n = 2n - 1, we obtain

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

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

m^{th} term of A.P/n^{th} term of A.P = a + (m - 1) d/a + (n - 1) d ....(3)

From (2) and (3) , we obtain

m^{th} term of A.P/n^{th} term of A.P = (2m - 1)/(2m - 1)

Hence, proved

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Question 12

## The ratio of the sums of m and n terms of an A.P. is m^{2} : n^{2} . Show that the ratio of m^{th} and n^{th} term is (2m – 1) : (2n – 1)

**Summary**:

The ratio of the sum of m and n terms of an A.P was given to be m^{2}/n^{2}. We have proved that the ratio of mth and the nth term is (2m - 1)/(2m - 1)