# How do you find the first term and common difference that has a sum of its first 10 terms equal to 250 and whose 6th term is 32?

Sep 13, 2015

The first term is $- 38$ (negative $38$).
The common difference is $14$.

#### Explanation:

So, we are talking about arithmetic progression, that is (this is the definition) a sequence of numbers, starting with some first one, $a$, and each consecutive one differed from the previous by the common difference $d$.

That is, we are talking about a sequence
$a , a + d , a + 2 \cdot d , a + 3 \cdot d , \ldots , a + N \cdot d , \ldots$

Assuming you don't remember the formula for this, let's derive a formula for a sum of the first $N$ terms of this sequence:
${S}_{N} = \left[a\right] + \left[a + d\right] + \left[a + 2 \cdot d\right] + \ldots + \left[a + \left(N - 2\right) \cdot d\right] + \left[a + \left(N - 1\right) \cdot d\right]$

Since the sum does not change if we change the order of summation, we can summarize it in opposite order:
${S}_{N} = \left[a + \left(N - 1\right) \cdot d\right] + \left[a + \left(N - 2\right) \cdot d\right] + \ldots + \left[a + 2 \cdot d\right] + \left[a + d\right] + \left[a\right]$

Sum both ${S}_{N}$ adding first term to first term, second term to second term etc., ${\left(N - 1\right)}^{t h}$ term to corresponding ${\left(N - 1\right)}^{t h}$ term, getting:

${S}_{N} + {S}_{N} =$
$= \left\{\left[a\right] + \left[a + \left(N - 1\right) d\right]\right\} +$
$+ \left\{\left[a + d\right] + \left[a + \left(N - 2\right) d\right]\right\} + \ldots$
$\ldots + \left\{\left[a + \left(N - 2\right) d\right] + \left[a + d\right]\right\} +$
$+ \left\{\left[a + \left(N - 1\right) d\right] + \left[a\right]\right\}$

Notice, that the results of summation in each $\left\{\ldots\right\}$ is the same, $2 a + \left(N - 1\right) \cdot d$.

Therefore,
$2 {S}_{N} = \left\{2 a + \left(N - 1\right) \cdot d\right\} \cdot N$

and
${S}_{N} = a \cdot N + \frac{\left(N - 1\right) \cdot N \cdot d}{2}$

The problem states that for $N = 10$ ${S}_{10} = 250$.
Therefore, we have one equation:
(Eq. 1) $250 = 10 a + 45 d$

Since ${6}^{t h}$ term is $32$, we have another equation:
$a + \left(6 - 1\right) d = 32$ or
(Eq. 2) $a + 5 d = 32$

We have a system of two equations, Eq. 1 and Eq. 2, with two unknowns $a$ and $d$.
It is easy to solve it using a substitution method.

From equation Eq. 2:
$a = 32 - 5 d$

Substitute this value for $a$ into equation Eq. 1:
$250 = 10 \left(32 - 5 d\right) + 45 d$
or
$250 = 320 - 50 d + 45 d$,
from which:
$5 d = 70$ and
$d = 14$
That is the common difference of our sequence.

Back to the unknown $a = 32 - 5 d$:
$a = 32 - 5 \cdot 14 = - 38$

Checking (ALWAYS RECOMMENDED)

The first 10 members of this sequence that starts with $- 38$ and adds $14$ to each new member are
$- 38 , - 24 , - 10 , 4 , 18 , 32 , 46 , 60 , 74 , 88$.
Their sum is indeed $250$ and their ${6}^{t h}$ term is indeed $32$.
Checking confirms the validity of the answer.