Find the first term of an arithmetic progression, given the 7th and the 16th terms

Question

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..

Answer 1

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

$$a+12=-1 \Rightarrow a=-13$$

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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.

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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

$$b=d \text{ and } a-d=c$$

Answer 2

Hint $ $ You're on the right track. The AP is essentially a discrete line, with slope being the common difference. Hence your formula for the difference is the usual slope formula for the line. Since the point $\,(\color{#c00}{1,y})\,$ lies on the line through $(\color{#0a0}{7,-1}),(\color{#90f}{16,17}),\,$ equating slopes yields

$$ \frac{\color{#c00}y-(\color{#0a0}{-1})}{\color{#c00}1-\color{#0a0}7}\,=\, \frac{\color{#90f}{17}-(\color{#0a0}{-1})}{\color{#90f}{16}-\color{#0a0}7}\, =\, 2\ \Rightarrow\ y = -13$$