Question

An arithmetic sequence has a common difference equal to 10 and its 6th term is equal to 52. Find its 15th term.​

Answer 1

Answer:

a₁₅ = 142

Step-by-step explanation:

General form of an arithmetic sequence:

$\boxed{a_n=a+(n-1)d}$

where:

• $a_n$ is the nth term.
• a is the first term.
• d is the common difference between terms.
• n is the position of the term.

Given:

• d = 10
• a₆ = 52

Substitute the given values into the formula and solve for a:

$\begin{aligned}a_n&=a+(n-1)d\\52&=a+(6-1)10\\52&=a+(5)10\\52&=a+50\\a&=52-50\\\implies a&=2\end{aligned}$

Substitute the found value of a and the given value of d into the general formula to create an equation for the nth term:

$\boxed{a_n=2+10(n-1)}$

To find the 15th term, substitute n = 15 into the found equation:

$\begin{aligned}a_n&=2+10(n-1)\\a_{15}&=2+10(15-1)\\a_{15}&=2+10(14)\\a_{15}&=2+140\\\implies a_{15}&=142\end{aligned}$

Therefore, the 15th term of the given arithmetic sequence is 142.

Answer 2

Answer: a₁₅=142

Step-by-step explanation:

$6th\ term\ is\ a_6=52\ \ \ \ \ \\a\ common \ difference\ d\ is\ 10\\to\ find\ 15th\ term\ a_{15}\\\\\boxed {a_n=a_1+d(n-1)}\\\\Hence,\\a_6=a_1+10(6-1)\\52=a_1+10(5)\\52=a_1+50\\52-50=a_1+50-50\\2=a_1\\Thus,\ a_1=2\\a_{15}=a_1+10(15-1)\\a_{15}=2+10(14)\\a_{15}=2+140\\a_{15}=142$