# 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

1. 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
2. a+4d = 3(a+d)
a+11d = 2(a+5d)+1

Solve for a and d, then

T16 = a+15d

posted by Steve

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