Advertisements
Advertisements
Question
The 5th term of an A.P. is thrice the second term, and the 12th term exceeds twice the 6th term by 1. Find the 16th term.
Sum
Advertisements
Solution
an = a + (n − 1)d
The 5th term (a5) is thrice the 2nd term (a2).
a + 4d = 3(a + d)
a + 4d = 3a + 3d
4d − 3d = 3a − a
d = 2a ...(1)
The 12th term (a12) exceeds twice the 6th term (a6) by 1.
a + 11d = 2(a + 5d) + 1
a + 11d = 2a + 10d) + 1
11d − 10d = 2a − a + 1
d = a + 1 ...(2)
Since both equations are equal to d, we can set them equal to each other:
2a = a + 1
a = 1
Now, substitute a = 1 into Equation 1.
d = 2a
d = 2 × 1
d = 2
an = a + (n − 1)d
a16 = 1 + (16 − 1)2
a16 = 1 + (15)2
a16 = 1 + 30
a16 = 31
shaalaa.com
Is there an error in this question or solution?
