maths ques on arithmetic sequences (1 Viewer)

[lilkiwifruit]

I know this question should be relatively easy but i just can't work it out ><"

The sum of the first 10 terms of an arithmetic series is 100 and the sum of the next 10 terms is 300. Find the series.

Thank you!

 

[SoulSearcher]

the sum of the first 10 terms of an artimatice sequence is as follows:

S₁₀ = 5 (2a + 9d)

the sum of the next ten sequences is equal to S₂₀ - S₁₀

S₂₀ = 10 (2a + 19d)

therefore S₂₀ - S₁₀ equals

10 (2a + 19d) - 5 (2a + 9d) = 20a + 190d - 10a - 45d = 10a + 145d

since S₁₀ = 100, then 5 (2a + 9d) = 100

and S₂₀ - S₁₀ = 300, then 10a + 145d = 300

Now 10a + 45d = 100 ... (1)
and 10a + 145d = 300 ... (2)

by simultaneous equations, where (2) - (1)

100d = 200, therefore d = 2

substituting d into 10a + 45d = 100,
10a + 90 = 100
10a = 10
a=1

therefore T_n = 1 + 2(n-1)

 

[pLuvia]

Damn 2u, we haven't even done series and stuff yet

 

[insert-username]

substituting d into 10a + 45d = 100,
10a + 90 = 100
10a = 10
a=5

Be careful.

Damn 2u, we haven't even done series and stuff yet

I recommend you do it yourself outside of class - it's not that difficult (apart from the loan repayments - that may need some explaining) - it's all relatively straightforward.

substituting d into 10a + 45d = 100,
10a + 90 = 100
10a = 10
a=5

Let d = common difference
Let a = first term
Let n = number of terms

S₁₀ = 100

S₂₀ = 400 (S₂₀ - S₁₀ = 300)

From equation 1: 5(2a + 9d) = 100 (sum of an arithmetic series)

So 10a + 45d = 100

So 10a = 100 - 45d (1)

From equation 2: 10(2a + 19d) = 400

20a + 190d = 400 (2)

From here, substitute (1) into (2), and solve the simultaneous equations. Answer below.

Spoiler

2(100 - 45d) + 190d = 400

200 - 90d + 190d = 400

200 + 100d = 400

100d = 200

d = 2

Therefore a = 1

Therefore the series is equal to 1 + 2(n-1)

I_F

 

[felixcthecat]

yea anyways!~

i assume u kno the AP equation T_n = n/2 [2a + (n-1)d]

S₁₀ = 10/2 (2a + 9d)------------(1)
S₂₀ = 20/2 (2a + 19d)----------(2)

to find only the sum of the "next 10 terms", which excludes the "first 10 terms":
S₂₀ - S₁₀
(2) - (1)
= 10(2a+19d) - 5(2a+9d)
= 10a + 145d

So: 10a + 45d = 100----------(a)
10a + 145d= 300----------(b)

(b)-(a)
100d= 200
d=2

plug 'd' into equation (a) [or b, you choose]
10a + 45(2) = 100
10a + 90 = 100
10a= 10
a= 1

the 'nth term' equation of an AP is: T_n = a + (n-1)d
so you know: d=2, a=1 .... so plug it into the 'nth term' equation
= 1 + (n-1)2
= 1 + 2n - 2
= -1 + 3n

hence if you put n=1,2,3... you'd get 1,3,5....

there we go~ all done ^0^

 

[lilkiwifruit]

Thank you for the help everyone!!! My answer finally matched the one in the textbook, 1,3,5,7 ~yay!