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How many terms of the A.P. 6 + 10 + 14 + ... has the sum 880? - Mathematics

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Question

How many terms of the A.P. 6 + 10 + 14 + ... has the sum 880?

Sum
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Solution

6 + 10 + 14 + ...

Sn = 880

a = 6

d = 10 − 6

= 4

Sn = `n/2[2a + (n - 1)d]`

880 = `n/2[2(6) + (n - 1)4]`

880 = `n/2[12 + (n - 1)4]`

880 = `n/2 xx 2 [6 + 2n - 2]`

880 = n[2n + 4]

880 = 2n2 + 4n

880 = 2(n2 + 2n)

`880/2 = n^2 + 2n`

440 = n2 + 2n

n2 + 2n − 440 = 0

n = `(-2 ± sqrt(4 + 4 (440)))/2`

= `(-2 ± sqrt(4 + 1760))/2`

= `(-2 ± sqrt(1764))/2`

= `(-2 ± sqrt((42)^2))/2`

= `(-2 ± 42)/2`

= `(-2 + 42)/2, (-2 - 42)/2`

= `40/2, (-44)/2`

n = 20

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Chapter 9: Arithmetic and geometric progression - Exercise 9C [Page 187]

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Nootan Mathematics [English] Class 10 ICSE
Chapter 9 Arithmetic and geometric progression
Exercise 9C | Q 4. (a) | Page 187
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