Question

The first term of an AP is 3 and the eleventh term is 18.find the number of terms in the progression if the sum is 81.

Answer 1

d = (T11-T1)/10 = 3/2

n/2 (2*3+(n-1)(3/2)) = 81
n = 9

Answer 2

n-th member in AP

an = a1 + ( n - 1 ) d

d = the difference between successive terms

In this case :

a11 = a1 + ( 11 - 1 ) d

a11 = a1 + 10 d = 18

18 = 3 + 10 d Subtract 3 to both sides

18 - 3 = 3 + 10 d - 3

15 = 10 d Divide both sides by 10

15 / 10 = 10 d / 10

1.5 = d

d = 1.5


The sum of the n terms of an arithmetic progression:

Sn = n [ 2 a1 + ( n - 1 ) d ] / 2

In this case

a1 = 3

d = 1.5

Sn = 81

so :

81 = n * [ 2 * 3 + ( n - 1 ) * 1.5 ] / 2

81 = n * [ 6 + 1.5 n - 1.5 ] / 2

162 = n * [ 6 + ( n - 1 ) * 1.5 ]

81 = n * ( 4.5 + 1.5 n ) / 2 Multiply both sides by 2

162 = n * ( 4.5 + 1.5 n )

162 = 4.5 n + 1.5 n ^ 2 Subtract 162 to both sides

162 - 162 = 4.5 n + 1.5 n ^ 2 -162

0 = 4.5 n + 1.5 n ^ 2 -162

1.5 n ^ 2+ 4.5 n -162 = 0


Solutions :

n = - 12

and

n = 9

Number of members can't be negative number so n = 9

Your AP :

3, 4.5, 6, 7.5 ,9, 10.5, 12, 13.5, 15