If a1, a2 ,a3 ........an are in Ap such that a5 +a7=16,then find a6.
In AP's difference between terms are equal, so
a5+x=a6
a7-x=a6
Adding the two equations cancels x, and expresses a6 in terms of the sum of a5 and a7.
Post your finding to check if you wish.
My answer a6=8.
Exactly, well done!
To find the value of a6, we need to use the properties of an arithmetic progression.
An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. In this case, let's assume that the common difference of the AP is d.
Given that a5 + a7 = 16, we need to find a6. Using the formula for the nth term of an AP, we can write:
a5 = a1 + (5-1)d
a7 = a1 + (7-1)d
Substituting these values into the equation a5 + a7 = 16, we have:
(a1 + 4d) + (a1 + 6d) = 16
2a1 + 10d = 16
a1 + 5d = 8 --(1)
Now, we want to find a6, which can be expressed as:
a6 = a1 + (6-1)d
a6 = a1 + 5d
We can substitute the value of a1 + 5d from equation (1), and we have:
a6 = 8
Therefore, the value of a6 is 8.