FIND The 6th And 5th Term Of The A.P Whose First Term Is 6 And Common Difference Is ?
You are expecting a miracle, aren't you?
sum of n terms of 1st = (n/2)(10 + 36(n-1))
= (n/2)(36n - 26)
sum of 2n terms of 2nd = (2n/2)(72 + 5(2n-1))
= n(10n + 67)
(n/2)(36n - 26) = n(10n + 67)
divide by n, and multiply by 2
36n - 26 = 20n + 134
16n = 160
n = 10
check:
sum(10 for 1st) = 5(10 + 36(9)) = 1670
sum(20 for 2nd) = 10(72 + 19(5)) = 1670
all is good!
Let the common difference be d.
The sum of the first four terms is
5 + (5+d) + (5+d+d) + (5+d+d+d)
= 20 + 6d
The sum of the next 4 terms is
5+4d + 5+5d + 5+6d + 5+7d = 20 + 22d
You have been told that
20 +6d = (1/2)(20 + 22d)= 10 + 11d
Therefore
5d = 10
d = 2
Check: does
5+7+9+11 = (1/2)(13+15+17+19) ?
32 = (1/2)(64)
To find the 6th and 5th terms of an arithmetic progression (A.P.), we need to know the common difference.
In the question, you mentioned that the first term of the A.P. is 6, but you didn't provide the common difference. Please provide the common difference so that I can help you find the 6th and 5th terms of the A.P.