Find the first term of an arithmetic progression, given the 7th and the 16th terms
Matthew Harrington 
I tried to solve this problem by first finding out the common difference by using the formula $$ \text{common difference} = \frac{T_p-T_q}{p-q} $$ with $T_7=-1$ and $T_{16}=17$. But now I'm not able to find the first term. I tried out many methods. Please help me out..
$\endgroup$ 42 Answers
$\begingroup$For an AP, the $n^{th}$ term is given by:
$$a_n=a+(n-1)d$$
,where $a$ is the first term and $d$ is the common difference
In this case
$$a_7=a+6d=-1\tag{1}$$
$$a_{16}=a+15d=17\tag{2}$$
Subtracting $(1)$ from $(2)$, we get
$9d=18 \Rightarrow d=2\tag{3}$
Now, plugging $(3)$ in $(1)$, we get
$$a+12=-1 \Rightarrow a=-13$$
An alternative (but similar) approach:
It is not hard to see that an AP behaves like a linear function.(perhaps from it's definition)
Let the general term of the AP be $$a_n=bn+c$$, where $b$ and $c$ are some constants.
Using this we get
$$a_7=7b+c=-1\tag{1}$$
$$a_{16}=16b+c=17\tag{2}$$
Subtracting $(1)$ from $(2)$, we get
$$9b=18\Rightarrow b=2\tag{3}$$
Using $(3)$ in $(1)$, we get
$$14+c=-1 \Rightarrow c=-15$$
Note:
Geometric version of this is pointed out by Bill Dubuque in a comment.
Using the two approaches we can observe that
$$a+(n-1)d=dn+(a-d)=bn+c$$
Since this an equation in $n$, for it to be an identity the coefficients must be same.
Comparing like power coefficients, we get
$$b=d \text{ and } a-d=c$$
$\endgroup$ $\begingroup$Hint $\ $ You're on the right track. The AP is essentially a discrete line, with slope being the common difference. So your formula for the difference is the usual slope formula for the line. Since $\,(1,y)\,$ lies on the line through $(7,-1),(16,17),\,$ equating slopes we find
$$ \frac{y-(-1)}{1-7}\,=\, \frac{17-(-1)}{16-7}\, =\, 2\ \Rightarrow\ y = -13$$
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