Another arithmetic progression

Relevant wiki: Arithmetic Progressions

Use the formula $a_n=a_m+(n-m)d$ to compute for $d$ . We have

$a_{19}=a_{10}+(n-m)(d)$

$76=40+(19-10)(d)$

$d=4$

So, the fourth term ( $a_4$ ) is:

$a_4=a_{10}+(4-10)(d)=40+(4-10)(4)=16$

or

$a_4=a_{19}+(4-19)(d)=76+(4-19)(4)=16$

Alternate solution:(compute the first term)

Use the formula $a_n=a_1+(n-1)(d)$ . We have

$a_{10}=a_1+(n-1)(d)=a_1+9d$ or $40=a_1+9d$ $\color{#D61F06}(1)$

$a_{19}=a_1+(n-1)(d)=a_1+18d$ or $76=a_1+18d$ $\color{#D61F06}(2)$

From $\color{#D61F06}(1)$ and $\color{#D61F06}(2)$ , we get $a_1=4$ and $d=4$ .

So $a_4=a_1+(n-1)(d)=4+3(4)=16$

nth term is a+(n-1)d form equations find a and d then ans is 16