Arithmetic Series Help

[SabziiKumari]

An AP has a common difference of 3.
Given the nth term is 32.
The sum of the first n terms is 185
calculate the value n

i do not understand how i would work this out.

 

[soroban]

Hello, SabziiKumari!

An AP has a common difference of 3.
Given the nth term is 32.
The sum of the first n terms is 185
Calculate the value of n

You're expected to know the formulas for an A.P.

. . \(\displaystyle \begin{Bmatrix}a &=& \text{first term} \\ d &=& \text{common difference} \\ n &=& \text{number of terms}\end{Bmatrix}\)


\(\displaystyle n^{th}\text{ term: }\;a_n \:=\:a + (n-1)d\)

\(\displaystyle \text{We are told: }\:a_n \,=\,32\!:\quad a + 3(n-1) \:=\:32 \quad\Rightarrow\quad a \:=\:35-3n\quad{\bf[1]}\)


\(\displaystyle \text{Sum of first }n\text{ terms: }\:S_n \:=\:\tfrac{n}{2}\bigg[a + (n-1)d\bigg]\)

\(\displaystyle \text{We are told: }\:S_n \,=\,32\!:\quad \tfrac{n}{2}[a + 3(n-1)] \:=\:185 \quad\Rightarrow\quad 3n^2 + 2na - 3n \:=\:370\quad{\bf[2]}\)


\(\displaystyle \text{Substitute {\bf[1]} into {\bf[2]}: }\;3n^2 + 2n(35-3n) - 3n \:=\:370 \quad\Rightarrow\quad 3n^2 - 67n + 370 \:=\:0\)

\(\displaystyle \text{Factor: }\n-10)(3n-37) \:=\:0 \quad\Rightarrow\quad n \:=\:10,\:\tfrac{37}{3}\)


. . \(\displaystyle \text{Therefore: }\:n \,=\,10\)

 

[SabziiKumari]

thank you

i do not understand how you came about the [2]

where did the 2na -3n come from ?

 

[mmm4444bot]

Multiply both sides of the equation by 2, to clear the fraction n/2. Then expand and combine like-terms.

 

[SabziiKumari]

thank you

but if you multiplied both sides of equation by 2, wouldn't the 3(n-1) be 6n-6 if you multiply by 2 ?
that is what im stuck on .

 

[mmm4444bot]

Okay, I see what you're thinking.

When you multiply the lefthand side by 2, it cancels with the 2 in the denominator. So, the 3(n - 1) is not multiplied by 2.

 

[SabziiKumari]

ok
thats fine

but then how did the 2na come about ?
cause if you timed out the n with the equation would you not get na not 2na ?

 

[mmm4444bot]

You're right. There's a factor of 2 missing.

It should be 2a.

(n/2)*[2a + 3(n - 1)] = 185

Sorry, I missed the typo.

 

[mmm4444bot]

\(\displaystyle \text{Sum of first }n\text{ terms: }\:S_n \:=\:\tfrac{n}{2}\bigg[2a + (n-1)d\bigg]\)

\(\displaystyle \text{We are told: }\:S_n \,=\,32\!:\quad \tfrac{n}{2}[2a + 3(n-1)] \:=\:185 \quad\Rightarrow\quad 3n^2 + 2na - 3n \:=\:370\quad{\bf[2]}\)

 

[SabziiKumari]

ohh
yes it makes sense now !
thank you very much.

 

[SabziiKumari]

also i have posted another thread with 2 questions that i need help on, if you could help me on those
thanks