# 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

1. 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

2. 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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