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Find the sum given below :

7 + 10 + 13 + ... + 46

 

 

 

 
 
 
 
 

Answers (1)

Given AP is :

7 + 10 + 13 + ... + 46

Here a=7, d=3, a_n=46

a_n=a+(n-1)d

46=7+(n-1)3

39=3(n-1)

n-1 = 13

n = 14

S_n=\frac{n}{2}[2a+(n-1)d]

S_n=\frac{14}{2}[2 \times 7+(14-1) \times 3]

S_n=7 [14+39]

S_n=7 \times 53

S_n=371

Posted by

Ravindra Pindel

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