The 11th term of an arithmetic profression is 1. The sum of the first 10 terms is 120. Find the 4th term.
Hi Tesphen.
The sum of the first 11 terms is 121. Hence the 6th term in the series is $\displaystyle \frac{121}{11}=11.$ (NB: The sum of an odd number of terms of an A.P. is the product of the middle term and the number of terms.) The common difference is therefore $\displaystyle \frac{\mathrm{11th\ term - 6th\ term}}{11-6}=\frac{1-11}5=-2.$ You should be able to find the 4th term from here.
$\displaystyle
S_n = \tfrac{1}
{2}n(a_1 + a_n ) = \frac{1}
{2}11(a_1 + 1) \Rightarrow a_1 = 21
$
$\displaystyle
a_1 + a_n = 2a_1 + d(n - 1) = 42 + 10d \Rightarrow d = - 2
$
$\displaystyle
\therefore \sum\limits_{k = 1}^4 {a_1 + (k - 1)d} = 21 + 19 + 17 + 15 = 72
$
4th term is 15. Sorry, somehow I thought the question was asking for the sum up to the 4th term.
Hello, Tesphen!
I must assume you know these formulas:
. . . . . . . . . $\displaystyle n^{th}\text{ term: }\;a_n \:=\:a_1 + (n-1)d$
$\displaystyle \text{Sum of the first }n\text{ terms: }\:S_n \:=\:\frac{n}{2}\left[2a_1 + (n-1)d\right]$
where: .$\displaystyle a_1$ = first term, $\displaystyle d$ = common difference.
The 11th term of an arithmetic profression is 1.
The sum of the first 10 terms is 120.
Find the 4th term.
The 11th term is 1: .$\displaystyle a_{11} \:=\:a_1 + 10d \:=\:1$ .[1]
The sum of the first 10 terms is 120:
. . $\displaystyle S_{10} \:=\:\frac{10}{2}(2a_1 + 9d) \:=\:120 \quad\Rightarrow\quad 2a_1 + 9d \:=\:24 $ .[2]
Solve the system: .$\displaystyle \begin{array}{cccc}{\color{blue}[1]} & a_1+10d &=& 1 \\
{\color{blue}[2]} & 2a_1 + 9d &=& 24 \end{array}\quad\Rightarrow\quad\begin{Bmatrix} a_1 &=& 21 \\ d &=& \text{-}2\end{Bmatrix}$
$\displaystyle \text{Therefore, the sequence is: }\;21, 19, 17, 15, 13, \hdots$
. . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \uparrow$
. . . . . . . . . . . . . . . . . . . . . $\displaystyle ^{\text{4th term}}$