Calculate the sum of the first 15 terms of an arithmetic progression is the one where a3= 1 and a7=7. (I need the procedure plea
1 answer:
Answer:
S₁₅ = 127.5
Step-by-step explanation:
the nth term of an arithmetic progression is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
given a₃ = 1 and a₇ = 7 , then
a₁ + 2d = 1 → (1)
a₁ + 6d = 7 → (2)
subtract (1) from (2) term by term to eliminate a₁
0 + 4d = 6
4d = 6 ( divide both sides by 4 )
d = 1.5
substitute d = 1.5 into (1) and solve for a₁
a₁ + 2(1.5) = 1
a₁ + 3 = 1 ( subtract 3 from both sides )
a₁ = - 2
the sum to n terms of an arithmetic progression is
=
[ 2a₁ + (n - 1)d ]
with a₁ = - 2 and d = 1.5 , then
S₁₅ = [ (2 × - 2) + (14 × 1.5) ]
= 7.5(- 4 + 21)
= 7.5 × 17
= 127.5
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