Question

Please helpIf the 100th term of an arithmetic sequence is 595, and its common difference is 6, thenits first term a1= ,its second term a2= ,its third term a3=

Answer 1

Given

100th term of an arithmetic sequence is 595 and common difference , d = 6

Find

First three terms of arithmetic sequences.

Explanation

As we know the general nth term of an arithmetic sequence is given by

$a_n=a+(n-1)d$

we have given 100th term = 595 , so

$\begin{gathered} a_{100}=a+(100-1)6 \\ 595=a+99\times6 \\ 595-594=a \\ a=1 \end{gathered}$

so , first term = 1

second term = a + 6 = 7

third term = a + 2d = 1 +2*6 = 13

Final Answer

Therefore , the first terms of an arithmetic sequences are

$a_1=1,a_2=7,a_3=13$