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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How many terms of the series 3 + 7 + 11 + 15 + .... must be taken to make the sum 820?

Answer 1

Solution

a=3

d=7-3=4

let the number of terms is n

now...

s(n)=[n(2×3+(n-1))]/2

$820 \times 2 = n(6 + 4n - 4) \\ = > 1640 =2n + 4n {}^{2} \\ = > 2n {}^{2} + n - 820 = 0 \\ = > 2n {}^{2} + 41n - 40n - 820 = 0 \\ = > n(2n + 41) - 20(2n + 41) = 0 \\ = > (2n + 41)(n - 20) = 0$

taking only positive value of n(cause n is a natural number)

n=20

so the number of terms is 20

hope this helps you........

Answer 2

Hey!!

Solution

a=3

d=7-3=4

let the number of terms is n now...

s(n)=[n(2×3+(n-1))]/2

820 × 2 =n (6+4n-4)

1640 = 2n +4n^2

2n^2+n-820=0

2n^2+41n-40n-820=0

n(2n+41)-20(2n+41)=0

(2n+41) (n-20)=0

taking only positive value of n(cause n is a natural number)n=20

So, the number of terms is 20.

Hope it helps you... ☺️✌️

$\huge\bold\pink{Thanks}$