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 the sum of the first 15 terms of an arithmetic sequence is 345. If a5=65, find a1 and d

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Explanation:

the sum of the first n terms of an arithmetic sequence is

Sn = n/2 × (a1 + an)

which is equivalent to

Sn = n/2 × (2a1 + (n-1)×d)

d is the constant, common difference between consecutive terms.

a1 is the first term.

an is the last term of the summed sequence.

so, we know

S15 = 345 = 15/2 × (2a1 + 14d) = 15×(a1 + 7d)

23 = a1 + 7d

a5 = 65 = a1 + 4d

remember,

a1 = a1

a2 = a1 + d

a3 = a2 + d = a1 + d + d = a1 + 2d

...

that means we have 2 equations with 2 variables (a1 and d) :

23 = a1 + 7d

65 = a1 + 4d

let's simply subtract the second from the first equation :

23 = a1 + 7d

- 65 = a1 + 4d

---------------------

- 42 = 0 + 3d

3d = -42

d = -14

and now we use that in one of the original equations like

23 = a1 + 7×-14 = a1 - 98

a1 = 121

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