Maths
The 5th term of an arithmetic progression is 3times of 2nd term and 12th term exceeds 2times of 6th term by 1. Find the 16th term

a5 = 3 a2
a12 = 2 a6 + 1
For AP:
an = a1 + ( n  1 ) d
a1 = initial term of an arithmetic progression
d = common difference of successive members
a2 = a1 + ( 2  1 ) d
a2 = a1 + 1 * d
a2 = a1 + d
a5 = a1 + ( 5  1 ) d
a5 = a1 + 4 d
a6 = a1 + ( 6  1 ) d
a6 = a1 + 5 d
a12 = a1 + ( 12  1 ) d
a12 = a1 + 11 d
Now:
a5 = 3 a2
a5 = 3 ( a1 + d )
a5 = 3 a1 + 3 d
a5 = a5
a1 + 4 d = 3 a1 + 3 d Subtract a1 to both sides
a1 + 4 d  a1 = 3 a1 + 3 d  a1
4 d = 2 a1 + 3 d Subtract 3 d to both sides
4 d  3 d= 2 a1 + 3 d  3 d
d = 2 a1
a12 = 2 a6 + 1
a1 + 11 d = 2 ( a1 + 5 d ) + 1
a1 + 11 d = 2 a1 + 10 d + 1 Subtract a1 to both sides
a1 + 11 d  a1 = 2 a1 + 10 d + 1  a1
11 d = a1 + 10 d + 1 Subtract 10 d to both sides
11 d  10 d = a1 + 10 d + 1  10 d
d = a1 + 1
d = d
2 a1 = a1 + 1 Subtract 1 to both sides
2 a1  1 = a1 + 1  1
2a1  1 = a1 Subtract a1 to both sides
2a1  1  a1 = a1  a1
a1  1 = 0 Add 1 to both sides
a1  1 + 1 = 0 + 1
a1 = 1
d = 2 a1
d = 2 * 1
d = 2
an = a1 + ( n  1 ) d
a16 = a1 + ( 16  1 ) d
a16 = a1 + 15 d
a16 = 1 + 15 * 2
a16 = 1 + 30
a16 = 31
posted by Bosnian

a+4d = 3(a+d)
a+11d = 2(a+5d)+1
Solve for a and d, then
T16 = a+15dposted by Steve
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