Let sum of n terms of a series be n (2n–1). Find its mth terms.

 

Let S_m and S_m–1 denote the sum of first m and (m – 1) terms respectively.

S_m = T₁ + T₂ + T₃ + ……. + T_m–1 + T_m

S_m = T₁ + T₂ + T₃ + ……. + T_m–1

Subtracting

S_m – S_m–1 = T_m

⇒ T_m = (m(2m–1))–(m–1)(2(m–1)–1))

        = (2m² – m)–(2m² – 5m + 3)

        = 4m – 3

 

 

        The sum of n terms of two A.P.’s are in the ratio 3n + 2: 2n+3. Find  the ratio of their 10^th terms.

 

        Let a, a + d, a + 2d, a + 3d, ……………

        A, A + D, A + 2D, A + 3D, ………………

        be two A.P.’s

            n/2[2a+(n–1)d]/n/2[2A+(n–1)d] = 3n+2/2n+3       (given)

        ⇒ a+n–1/2d/A+n–1/2D = 3n+2/2n+3

        ⇒ To get the ratio of 10^th terms put n–1/2 = 9