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