The sum of third and seventh term of an A.P.is 6 and their product is 8.find the sum of 16th term of the A.p.
Answers
we know that Nth term is a + (n - 1)d
3rd term = a + (3 - 1)d = a + 2d
7th term = a + (7 - 1)d = a + 6d
sum of 3rd and 7th is 6
[a + 2d] + [a + 6d] = 6
a 2d + a + 6d = 6
2a + 8d = 6
2(a + 4d) = 6
a + 4d = 6/2 = 3
a = 3 - 4d .........(i)
their product = 8
[a + 2d]*[a + 6d] = 8
put the value of x from. (i)
[3 - 4d + 2d]*[3 - 4d + 6d] = 8
[3 - 2d]*[3 + 2d] = 8
3² - (2d)² = 8 ........[: (a+b)(a-b) =a²-b²]
9 - 4d² = 8
9 - 4d² - 8 = 0
9 - 8 - 4d² = 0
1 - 4d² = 0
1 = 4d²
1/4 = d²
d = +-(1/2)
put the value of d in (i)
a = 3 - 4d = 3 - 4(1/2) = 3 - 2 = 1
a = 3 - 4d = 3 -4(-1/2) = 3 -(-2) = 3+2 = 5
Hence,
according to a = 1,d = 1/2
16th term = a + 15d = 1 + 15(1/2) = 17/2
according to a = 5, d = -1/2
16th term = a + 15d = 5 + 15(-1/2)
= 5 - 7.5 = -2.5 = -5/2