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Illustration:
Let sum of n terms of a series be n (2n–1). Find its mth terms.
Solution:
Let Sm and Sm–1 denote the sum of first m and (m – 1) terms respectively.
Sm = T1 + T2 + T3 + ……. + Tm–1 + Tm
Sm = T1 + T2 + T3 + ……. + Tm–1
Subtracting
Sm – Sm–1 = Tm
⇒ Tm = (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 10th 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 10th terms put n–1/2 = 9
or n = 19
⇒ a+9d/A+9D = 3(19)+2/2(19)+3 = 59/41
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