Example 2 : Determine the AP whose 3rd term is 5 and the 7th term is 9 . Solution : a3 = a + ( 3 – 1 ) d = a + 2d = 5 --( i ) a7 = a + ( 7 – 1 ) d = a + 6d = 9 --( ii ) Solving the pair of linear equations ( i ) and ( ii ), we get a = 3 , d = 1 Hence , the required AP is 3 , 4 , 5 , 6 , 7 . . .
Example 3 : Check whether 301 is a term of the list of numbers 5 , 11 , 17 , 23 , . . .
Solution : d = a2 – a = 6 & a = 5 Let 301 be a term , say , the nth term of this AP . an = a + ( n – 1 ) d or 301 = 5 + ( n – 1 ) × 6 301 = 6n – 1 Or n = 302 / 6 which is a fraction . Thus 301 is not in the series .
Sum of first n terms of Arithmetic Progression
Suppose a person gets a salary of 1000 Rs & an increment of 100Rs every month . We want to know how much salary the person has drawn in 10 months . For this we need to find sum of first 10 terms of this AP with a = 1000 & d = 100 .
sum of the first n terms of an AP is given by S = n / 2 [ 2a + ( n – 1 ) d ] We can also write this as S = n / 2 [ a + a + ( n – 1 ) d ] or S = n / 2 ( a + an ) Sum of first n positive integers is given by Sn = n ( n + 1 )/ 2