Find the middle terms of the A.P. 7, 13, 19, ....,241.

Given,
Arithmetic progression =

7, 13, 19, \dots \dots . . 241

Here,
First term

a = 7,

Last term

t_{n} = 241,

Common difference

d = 13 - 7 = 6

Then applying the formula

t_{n} = a + (n - 1) d

Put the value in this formula

241 = 7 + (n - 1) 6

\frac{241 - 7}{6} = n - 1

\frac{234}{6} = n - 1

39 + 1 = n;

n = 40

Therefore, middle term of series

\frac{n}{2} = \frac{40}{2} = 20

\frac{n}{2} + 1 = 21

Now calculate the 20th and 21st terms

t_{20} = 7 + (20 - 1) 6

t_{20} = 7 + 19 \times 6

t_{20} = 7 + 114 = 121

t_{21} = 121 + 6 = 127

Hence, the middle terms are

121 and 127 .

Option 3 is correct.

JEE