Arithmetic Sequences (1 Viewer)

[Leasha23]

For finding the sum of an arithmetic sequence, the formula is:

Sum = n/2 [2a + (n-1)d], when the first term, difference, and number of terms is known.

However, what is the variation of the formula when both the last and first term are known? Is it:

Sum = n/2 (2a + l) ?

Thanks...

 

[tempco]

n/2 (a + l)

 

[iambored]

yeah, careful with that, i always forget he 2 on the first one

 

[coroneos]

do you know your time payments and stuff applications of GP

 

[Leasha23]

Originally posted by coroneos
do you know your time payments and stuff applications of GP

Yep I worked them out today! They actually arent that hard once you get the pattern worked out... I just always got the simple formula for arithmetic progressions mixed up

 

[mercury]

HOW TO REMEMBER

just think for arithmetic sequence

sum = 1/2 * total number of terms * (first term + last term)

eg. if you have like 1, 3, 5, 7 or something
you could pair them off like 1+7 = 8, 3+5 =8
so you'll have in this case 4/2 = 2 pairs of (first term + last term)

As for the other one:

S = n/2 (2a + (n-1)d)

you just need to remember, that
T_n = a + (n-1)d

so sub in to the other formula :
S = n/2 (first term + last term)
= n/2 (a + l)
= n/2 (a + a + (n-1)d)
= n/2 (2a + (n-1)d)

 

[Leasha23]

Originally posted by mercury
HOW TO REMEMBER

just think for arithmetic sequence

sum = 1/2 * total number of terms * (first term + last term)

eg. if you have like 1, 3, 5, 7 or something
you could pair them off like 1+7 = 8, 3+5 =8
so you'll have in this case 4/2 = 2 pairs of (first term + last term)

As for the other one:

S = n/2 (2a + (n-1)d)

you just need to remember, that
T_n = a + (n-1)d

so sub in to the other formula :
S = n/2 (first term + last term)
= n/2 (a + l)
= n/2 (a + a + (n-1)d)
= n/2 (2a + (n-1)d)

Thanks mercury you're an angel!

 

[mercury]

pleasure