Question

Please help with step by step answers to solving the problem!

The sum of the first n terms of an arithmetic sequence with first term a1 and a common difference d is Sn. Determine the sum if each term is increased by 5.

Answer 1

sn=[n(a1+an)/2]

an=a1+(n-1)d

Find your a1; d is given.

Answer 2

If you increase each term by 5 it means you are adding 5 to every single term

Answer 3

Total number of terms = n so
you have added 5 ,n times
i.e You added 5n
sum is sn +5n

Answer 4

I'm still confused on how writing out the formula to figure out the sum.

Answer 5

I think you are missing some information. All you are given is d, but not a1. You cant really find an, or sn without more info

Answer 6

It's like proving the theory of arithmetic sequence and finding its sum.

Answer 7

n/2(2a1+(n-1)d) + 5n

Answer 8

how did you get the 2?

Answer 9

sn +5n =n/2(2a1+(n-1)d) + 5n
because sn =n/2(2a1+(n-1)d)

Answer 10

You wanna me to prove sn= n/2(2a1+(n-1)d) ?

Answer 11

No, I get it sorta now. If I have any other questions I'll ask.

Answer 12

Isn't the formula for the sum of arithmetic sequence supposed to be n(a1+an)/2?

Answer 13

yeah But l= a1+(n-1)d

Answer 14

formulas are in the first post

Answer 15

so d = 5?

Answer 16

yes

Answer 17

Nope

Answer 18

d= d

Answer 19

even if each term is increased by 5 ,the common difference won't change
you should change you first term .put a1 +5 at the place of a1

Answer 20

you will get n/2(2a1+(n-1)d) + 5n

Answer 21

Can you write it out step by step and show me?

Answer 22

Sure
my if my AP is a1 , a1+ d , a1+ 2d ,............................... a1+(n-1)d
sn=n/2(2a1+(n-1)d)
If I add 5 to every every number AP becomes
a1+5 , a1+d+5, a1+2d+5 ,................................ a1+(n-1)d +5
instead of a1 we'll put a1+5 because our 1st term is now a1+5
check d hast change yet
sn =n/2(2(a1+5)+(n-1)d) =n/2(2a1+(n-1)d) + 5n

Answer 23

Sure
my if my AP is a1 , a1+ d , a1+ 2d ,............................... a1+(n-1)d
sn=n/2(2a1+(n-1)d)
If I add 5 to every number, AP becomes
a1+5 , a1+d+5, a1+2d+5 ,................................ a1+(n-1)d +5
instead of a1 we'll put a1+5 because our 1st term is now a1+5
check d hasn't change yet
sn =n/2(2(a1+5)+(n-1)d) =n/2(2a1+(n-1)d) + 5n

Answer 24

Thanks.

Answer 25

You're welcome .... I'm always ready to help those who want my help