The 3rd and 6th terms of an arithmetic progression are 94 and 85 respectively. Find the value of the 15th term.

Guest Jun 18, 2022

#1**+1 **

Hi Guest!

Let the first term of the arithmetic progression be \(a_0\)

Let the common difference between two successive terms be \(d\)

Then, \(a_n=a_0+d(n-1)\)

The third term is when n=3:

\(a_3=a_0+2d\)

We are given that the third term is 94:

\(94=a_0+2d\) (Equation 1)

The 6th term is when n=6:

\(a_6=a_0+5d\)

Given that the 6th term is 85:

\(85=a_0+5d \) (Equation 2)

Subtract Equation 2 from Equation 1:

\(94-85=a_0+2d-a_0-5d \iff 9 = -3d \iff d=-3\)

Substituting in Equation 1:

\(94=a_0+2(-3) \iff a_0=100\)

The 15th term is when n=15:

\(a_{15}=a_0+14d=100+14(-3)=100-42=58\)

Hope this helped!

Guest Jun 18, 2022