#### notes

In the sequence 5, 8, 11, 14, . . . the difference between two consecutive terms is 3.

Hence, this sequence is an A.P.

Here the first term is 5. If 3 is added to 5 we get the second term 8. Similarly to find 100th term what should be done?

First term Second term Third term . . .

Number 5, 5 + 3 = 8 8 + 3 = 11 . . .

In this way reaching upto 100th term will be time consuming. Let’s see if we can find any formula for it.

Generally in the A.P. t1, t2, t3, . . . If first term is a and common difference is d,

t_{1}= a

t_{2}= t_{1}+ d = a + d = a + (2 - 1) d

t_{3}= t_{2}+ d = a + d + d = a + 2d = a + (3 - 1)d

t_{4}= t_{3}+ d = a + 2d + d = a + 3d = a +(4 - 1)d

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

Using the above formula we can find the 100th term of the A.P. 5, 8, 11, 14, . . .

Here a = 5 d = 3

tn = a +(n - 1)d

t_{100}= 5 +(100 - 1) × 3

= 5 + 99 × 3

= 5 + 297

t_{100} = 302

100^{th} tem of this A.P. is 302.