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Solved Examples Based on Harmonic mean Illustration: Find the 4th and 8th term of the series 6, 4, 3, …… Solution: Consider1/6, /14, 1/3, ...... ∞ Here T2 – T1 = T3 – T2 = 1/12 ⇒ 1/6, 1/4, 1/3 is an A.P. 4th term of this A.P. = 1/6 + 3 × 1/12 = 1/6 + 1/4 = 5/12, And the 8th term = 1/6 + 7 × 1/12 = 9/12. Hence the 8th term of the H.P. = 12/9 = 4/3 and the 4th term = 12/5. Illustration: If a, b, c are in H.P., show that a/b+c, b/c+a, c/a+b are also in H.P. Solution: Given that a, b, c are in H.P. ⇒ 1/a, 1/b, 1/c are in A.P. ⇒ a+b+c/a, a+b+c/b, a+b+c/c are in A.P. ⇒ 1 + b+c/a, 1 + c+a/b, 1 + a+b/c are in A.P. ⇒ b+c/a, c+a/b, a+b/c are in A.P. ⇒ a/b+c, b/c+a, c/a+b are in H.P. Some Important Results • 1 + 2 + 3 +…+ n = n/2(n + 1) (sum of first n natural numbers). • 12 + 22 + 32 +…+ n2 = n(n+1)(2n+1)/6 (sum of the squares of first n natural numbers). • 13 + 23 + 33 +…+ n3 = n2(n+1)2/4 = (1 + 2 + 3 +…+ n)2 (sum of the cubes of first n natural numbers). • (1 – x)–1 = 1 + x + x2 + x3 +… –1 < x < 1. • (1 – x)–2 = 1 + 2x + 3x2 +… –1 < x < 1. Illustration: Find the nth term and the sum of n terms of the series 1.2.4 + 2.3.5 + 3.4.6 +… Solution: rth term of the series = r(r+1).(r+3)=r3 + 4r2 + 3r So sum of n terms = Σnr=1 r3 + 4Σnr=1 r2 + 3Σnr=1 r = (n(n+1)/2)2 + 4 n(n+1)(2n+1)/6 + 3n(n+1)/2 = n(n+1)/12 {3n2 + 19n + 26}. Illustration: Find the sum of the series 1.n + 2(n–1) + 3.(n–2) +…+ n.1. Solution: The rth term of the series is tr = (1 + (r – 1).1)(n + (r–1)(–1)) = r(n – r + 1) = r(n + 1) – r2 ⇒ Sn= Σnr=1 tr Σnr=1 (n+1)r – Σnr=1 r2 = (n+1) n.(n+1)/2 – n(n+1)(2n+1)/6 = n.(n+1)/2 [n + 1 – 2n+1/3] = n(n+1)(n–2)/6. To read more, Buy study materials of Sequences and Series comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
Illustration:
Find the 4th and 8th term of the series 6, 4, 3, ……
Solution:
Consider1/6, /14, 1/3, ...... ∞
Here T2 – T1 = T3 – T2 = 1/12 ⇒ 1/6, 1/4, 1/3 is an A.P.
4th term of this A.P. = 1/6 + 3 × 1/12 = 1/6 + 1/4 = 5/12,
And the 8th term = 1/6 + 7 × 1/12 = 9/12.
Hence the 8th term of the H.P. = 12/9 = 4/3 and the 4th term = 12/5.
If a, b, c are in H.P., show that a/b+c, b/c+a, c/a+b are also in H.P.
Given that a, b, c are in H.P.
⇒ 1/a, 1/b, 1/c are in A.P.
⇒ a+b+c/a, a+b+c/b, a+b+c/c are in A.P.
⇒ 1 + b+c/a, 1 + c+a/b, 1 + a+b/c are in A.P.
⇒ b+c/a, c+a/b, a+b/c are in A.P.
⇒ a/b+c, b/c+a, c/a+b are in H.P.
Some Important Results
• 1 + 2 + 3 +…+ n = n/2(n + 1) (sum of first n natural numbers).
• 12 + 22 + 32 +…+ n2 = n(n+1)(2n+1)/6 (sum of the squares of first n natural numbers).
• 13 + 23 + 33 +…+ n3 = n2(n+1)2/4 = (1 + 2 + 3 +…+ n)2 (sum of the cubes of first n natural numbers).
• (1 – x)–1 = 1 + x + x2 + x3 +… –1 < x < 1.
• (1 – x)–2 = 1 + 2x + 3x2 +… –1 < x < 1.
Find the nth term and the sum of n terms of the series 1.2.4 + 2.3.5 + 3.4.6 +…
rth term of the series = r(r+1).(r+3)=r3 + 4r2 + 3r
So sum of n terms = Σnr=1 r3 + 4Σnr=1 r2 + 3Σnr=1 r
= (n(n+1)/2)2 + 4 n(n+1)(2n+1)/6 + 3n(n+1)/2 = n(n+1)/12 {3n2 + 19n + 26}.
Find the sum of the series 1.n + 2(n–1) + 3.(n–2) +…+ n.1.
The rth term of the series is
tr = (1 + (r – 1).1)(n + (r–1)(–1))
= r(n – r + 1) = r(n + 1) – r2
⇒ Sn= Σnr=1 tr Σnr=1 (n+1)r – Σnr=1 r2 = (n+1) n.(n+1)/2 – n(n+1)(2n+1)/6
= n.(n+1)/2 [n + 1 – 2n+1/3] = n(n+1)(n–2)/6.
To read more, Buy study materials of Sequences and Series comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
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