Question

the first arithmetic sequence is 6 and the common difference is 7 what is the first 10 terms?

Answer 1

Given: The first term in arithmetic sequence is 6 and common difference is 7

           i.e a = 6 and d = 7

To Find: The first 10 terms in the sequence

Solution:

The formula for calculating nth term in an arithmetic progression is

$a_{n} = a_{1} + (n - 1)d$ where $a_{n}$ = nth term and $a_{1}$ = first term

So accordingly since first term is already 6,

Second Term:

$a_{2} = a_{1} + 1 * 7 = 6 + 7 = 13$

Third Term:

$a_{3} = a_{1} + 2 * 7 = 6 + 14 = 20$

Fourth Term:

$a_{4} = a_{1} + 3 * 7 = 6 + 21 = 27$

Fifth Term:

$a_{5} = a_{1} + 4 * 7 = 6 + 28 = 34$

Sixth Term:

$a_{6} = a_{1} + 5 * 7 = 6 + 35 = 41$

Seventh Term:

$a_{7} = a_{1} + 6 * 7 = 6 + 42 = 48$

Eight Term:

$a_{8} = a_{1} + 7 * 7 = 6 + 49 = 55$

Ninth Term:

$a_{9} = a_{1} + 8 * 7 = 6 + 56 = 62$

Tenth Term:

$a_{10} = a_{1} + 9 * 7 = 6 + 63 = 69$

So accordingly the first 10 terms in the arithmetic progression are 6, 13, 20, 27, 34, 41, 48, 55, 62, 69