In an arithmetic sequence...

Question

the third term is 10 and the fifth term is 16.

 

a) find the common difference

 

b) find the first term

 

c) find the sum the first 20th term of the sequence

Answer 1

Hello, thank you for taking the time to post your question!

The underlying formula that you want to use here for an arithmetic sequence is

an = a1 + (n – 1) * d

so when we’re given a3 = 10 and a5 = 16, that allows us to solve for the common difference by taking

(16 – 10) / (5 – 3)

= 6 / 2

= 3

So 3 is the common difference.

From there then you can take one of the terms like a3 = 10 and go back to the original equation to solve for a1

10 = a1 + (3 – 1)3

10 = a1 + 6

4 = a1

Meaning that the first term in the sequence is 4

Finally then for the sum you can use the formula

Sn = n/2(2a1 + (n – 1)d)   for the sum of the first “n” terms of the arithmetic sequence

S20 = 20/2(2(4) + (20 – 1)3)

S20 = 10(8 + 19(3))

S20 = 10(8 + 57)

S20 = 10(65)

S20 = 650

meaning that the sum of the first 20 terms is 650

I hope that helps get you moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

in an arithmetic series, an = a0+ (n-1) *d

 

that means the nth term is equal to the first term plus n-1 times the common difference.

 

In this case we are told a3=10 and a5=16

 

so:  a3= a+ 2*d =10    => a0= 10-2d     

and      

      a5= a+4*d = 16     => a0= 16-4d

 

so 10-2d = 16-4d => 2d=6 => d=3

 

now you can plug this value of d into any of those equations we got for a0 and find a0.

for example a0= 10-2(3) = 4 so the first term is 4

 

the formula for the sum of the first n terms is:

 

Sn = (n/2)* [2a0+ (n-1)*d]

 

so S20 = (20/2) * [2*4 + (19)*(3)] = 10* [65] = 650