In an AP, sum of a3 and a4 is equal to 3 times the sum of a1 and a2,if a4=14,find Sn
Answers
Answer:
Sn = 2n^2
Step-by-step explanation:
let initial term of the AP = a1 =a
and the common difference = d
nth term = an = a+(n-1)d = (n-1)th term + d
guven, a4 = 14 = a3 + d
so, a3 = a4 - d = 14 - d
similarly, a2 = a3 - d = 14 - d - d = 14 - 2d
a1 = a2 - d = 14 - 2d - d = 14 - 3d
given, a3 + a4 = 3(a1 + a2)
=> (14 - d) + 14 = 3{(14 - 3d) + (14 - 2d)}
=> 28 - d = 3(28 - 5d) = 84 - 15d
=> 15d - d = 84 - 28
=> 14d = 56
=> d = 56/14 = 4
a = a1 = 14 - 3d = 14 -3×4 = 14 - 12 = 2
Sn = [n{2×a + (n-1)d}]/2
= [n{2×2 + (n-1)4}]/2
= [n{4 + 4n - 4}]/2
= [n{4n}]/2
= 2n^2
Answer:
Step-by-step explanation:
Answer:
Sn = 2n^2
Step-by-step explanation:
let initial term of the AP = a1 =a
and the common difference = d
nth term = an = a+(n-1)d = (n-1)th term + d
guven, a4 = 14 = a3 + d
so, a3 = a4 - d = 14 - d
similarly, a2 = a3 - d = 14 - d - d = 14 - 2d
a1 = a2 - d = 14 - 2d - d = 14 - 3d
given, a3 + a4 = 3(a1 + a2)
=> (14 - d) + 14 = 3{(14 - 3d) + (14 - 2d)}
=> 28 - d = 3(28 - 5d) = 84 - 15d
=> 15d - d = 84 - 28
=> 14d = 56
=> d = 56/14 = 4
a = a1 = 14 - 3d = 14 -3×4 = 14 - 12 = 2
Sn = [n{2×a + (n-1)d}]/2
= [n{2×2 + (n-1)4}]/2
= [n{4 + 4n - 4}]/2
= [n{4n}]/2
= 2n^2
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