Maths
posted by Anonymous
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

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