Given a1=−6 and a2=−8. If the recurrence relation satisfies the equation an+1=2an+2+3an, then the value of a3+2a4 is ...
Explanation
We are given the first two terms of the sequence, a1=−6 and a2=−8, and the recursive equation:
an+1=2an+2+3an
Our goal is to find the value of a3+2a4. We will find the values of a3 and 2a4 step by step.
Finding the Third Term
We use the recursive equation by substituting n=1:
a1+1=2a1+2+3a1
a2=2a3+3a1
Substitute the known values of a1 and a2:
−8=2a3+3(−6)
−8=2a3−18
Move −18 to the left side:
2a3=−8+18
2a3=10
a3=5
Finding Two Times the Fourth Term
Next, we use the recursive equation again with n=2 to introduce the term a4:
a2+1=2a2+2+3a2
a3=2a4+3a2
Substitute the values a3=5 and a2=−8:
5=2a4+3(−8)
5=2a4−24
We can directly solve for 2a4 (since the question asks for a3+2a4):
2a4=5+24
2a4=29
Final Result
Now we sum the obtained values:
a3+2a4=5+29=34
Thus, the value of a3+2a4 is 34.