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