Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms li

Question

Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is:

Accepted Answer

Given: The A.P. has positive integer terms. The sum of the first three terms is $54$ , and the sum of the first twenty terms lies between $1600$ and $1800$ .

Find: The $11$ th term.

Let the A.P. be $a, a+d, a+2d, \dots$

From the sum of the first three terms,

S_3 = a + (a+d) + (a+2d) = 3a+3d = 54

So,

a+d = 18

Now use the sum of the first twenty terms:

S_{20} = \frac{20}{2}\left(2a+19d\right) = 10\left(2a+19d\right)

Given,

1600 < S_{20} < 1800

Therefore,

1600 < 10\left(2a+19d\right) < 1800

Dividing by $10$ ,

160 < 2a+19d < 180

Using $a = 18-d$ from $a+d=18$ ,

160 < 2(18-d)+19d < 180

160 < 36+17d < 180

124 < 17d < 144

\frac{124}{17} < d < \frac{144}{17}

So,

7.29 < d < 8.47

Since the A.P. is of positive integers, $d$ is an integer. Hence,

d = 8

Then,

a = 18-8 = 10

The $11$ th term is

a_{11} = a + 10d = 10 + 10\cdot 8 = 90

Therefore, the $11$ th term is $90$ . The correct option is C.

The solution marks option D, but its own working gives $90$ , which matches option C.

Given: The sum of the first three terms is $54$ .

Find: The $11$ th term using the relation among the first three terms.

In an A.P., the sum of the first three terms is three times the middle term. Hence the second term is

\frac{54}{3} = 18

So,

a+d = 18

Again, for the sum of the first twenty terms,

1600 < 10(2a+19d) < 1800

which gives

160 < 2a+19d < 180

Replacing $a$ by $18-d$ ,

160 < 36+17d < 180

This gives

d=8

and hence

a=10

Now compute the $11$ th term:

a_{11} = a+10d = 10+80 = 90

Therefore, the correct answer is $90$ , that is, option C.

Answer

Step-by-step solution

Common mistakes

Practice more Arithmetic Progression (AP) questions

Related questions