Question

Find the 11th term of AP from the end: 3, 8, 13, 18,….., 318​

Answer 1

Step-by-step explanation:

a= 3, d= 8-3=5

a11= a +(n-1)d

=3+(11-1)5

=3+50

=53

Answer 2

Required answer

268

Solution

Let's change the order from the end to the start.

Let the AP be defined by $a_{n}=318-5(n-1)$.

$\rightarrow a_{n}=318-5n+5$

$\rightarrow a_{n}=323-5n$

The 11th term is found by substitution of $n=11$

$\rightarrow a_{11}=323-55$

$\rightarrow a_{11}=268$

More information

• Sequence

Numbers organized in a rule. However, this is considered a function of the natural domain.

• Arithmetic Progression

A sequence that adds always equally.

• Geometric Progression

A sequence that multiplies always equally.

• Harmonic Progression

A sequence of the inverse of AP.