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,
t2= t1+ d = a + d = a + (2 - 1) d
t3= t2+ d = a + d + d = a + 2d = a + (3 - 1)d
t4= t3+ d = a + 2d + d = a + 3d = a +(4 - 1)d
We get tn= 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
t100= 5 +(100 - 1) × 3
= 5 + 99 × 3
= 5 + 297
t100 = 302
100th tem of this A.P. is 302.
Six year before, the age of mother was equal to the square of her son's age. Three year hence, her age will be thrice the age of her son then. Find the present ages of the mother and son.