## Intermediate Algebra (6th Edition)

$a_{25}= 75$
$a_{n}$ of the arithmetic sequence is $a_{n} = a_{1} + (n-1)d$ Similarly, Second term of arithmetic sequence is $a_{2} =6$ $a_{2} = a_{1} + (2-1)d$ $a_{2} = a_{1} + d$ $a_{1} + d =6$ Equation $(1)$ Tenth term of arithmetic sequence is $a_{10} =30$ $a_{10} = a_{1} + (10-1)d$ $a_{10} = a_{1} + 9d$ $a_{1} + 9d =30$ Equation $(2)$ Subtract Equation $(1)$ from Equation $(2)$ $a_{1} + 9d - (a_{1} + d )=30 - 6$ $a_{1} + 9d - a_{1} - d =24$ $8d = 24$ $d = 3$ Substituting $d$ value in Equation $(1)$ $a_{1} + d =6$ $a_{1} + 3 =6$ $a_{1}= 3$ Using $a_{1}$ , $d$ values and $n=25$ , $a_{25} = a_{1} + (25-1)d$ $a_{25} = 3 + (24)3$ $a_{25} = 3 +72$ $a_{25} = 75$