Question

find the sum of first 20 terms of the ap whose nth term is 3n-5

Answer 1

We are Given an=3n-5
                    So, a1=3(1)-5
                          a1=3-5= -2
                           
                          a2=3(2)-5
                          a2=1
              
                       So, d= a2-a1= 1-(-2)
                             d=3

                    So, S20= 20/2[2a+19d]
                          S20=10[2(-2)+19(3)]
                                =10[57-4]
                                =10*53
                                =530   
              So, Required Sum , S20= 530

Answer 2

Concept

Arithmetic is a series in which there is a common difference between the terms.

Given

nth term of an AP is 3n-5

Find

We are asked to find the sum of first 20 terms of the AP whose nth term is 3n-5

Solution

By considering given nth term

$a_{n}=3n-5$

We will find first three terms of an AP by putting 1,2 and 3 in nth term.

$a_1=3(1)-5=-2\\a_2=3(2)-5=1\\a_3=3(3)-5=4$

Now we can find the common difference by subtracting first term from second term.

$d=4-1=3$

Now we will find the sum by using formula

$S_n= \frac{n}{2}( 2a+(n-1)d)\\ S_20=\frac{20}{2}(2(-2)+(20-1)3)\\ S_20=10(-4+57)\\ S_20=10(53)\\S_20=530$

Hence the sum of first 20 terms of the AP whose nth term is 3n-5 is 530