Question

║⊕QUESTION⊕║
Mathematics is the most beautiful and most powerful creation of the human spirit
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CLASS 11
SEQUENCES AND SERIES
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If the sum of series 2, 5, 8, 11, ... is 60100, then find the value of n.

Answer 1

Answer:

60100 = n/2 (4+(n-1)3)

120200 = n(3n+1)

3n² + n - 120200 = 0

3n² + 601n - 600n - 120200 = 0

3n² + 601n - 600n - 120200 = 0n(3n + 601) - 200 (3n + 601) = 0

3n² + 601n - 600n - 120200 = 0n(3n + 601) - 200 (3n + 601) = 0(n - 200) (3n + 601) = 0

n = 200 (Ans.)

Answer 2

$\large \underline{ \underline{ \sf \: Solution : \: \: \: }}$

Given ,

$\star$First term = 2

$\star$Common difference = 3

$\star$Sum of nth term = 60100

We know that ,

$\large \sf \fbox{\fbox{S_{n} = \frac{n}{2} \bigg(2a + (n - 1)d \bigg)}}$

$\sf \implies 60100 = \frac{n}{2} (2 \times 2 + (n - 1)3) \\ \\ \sf \implies 120200 = n(4 + 3n - 3) \\ \\ \sf \implies 120200 = n(1 + 3n) \\ \\ \sf \implies 120200 = n + 3 {n}^{2} \\ \\ \sf \implies 3 {n}^{2} + n + 120200 = 0\\ \\ \sf \implies 3 {n}^{2} + 601n - 600n - 120200 = 0 \\ \\ \sf \implies</p><p>n(3n + 601) - 200 (3n + 601) = 0 \\ \\ \sf \implies </p><p>(n - 200) (3n + 601) = 0 \\ \\ \sf \implies </p><p>n = 200 \: \: or \: \: n = - \frac{601}{3} \: \: \{ignore \: negative \: value \: of \: n \}$

$\therefore$The required value of n is 200