Find the sum of the first 24 terms of the list of numbers whose n th term is given by an=3+2n
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Hey mate!
Since it is given that the nth teem is given by an = 3+2n,
a1 = 3+2(1) => 3+2 => 5
a2 = 3+2(2) => 3+4 => 7
a3 = 3+2(3) => 3+6 =>9
Hence, a = 5; d = 7-5 => 2; n = 24
Now, we know that,
Sn = n/2[2a+(n-1)d]
So,
S24 = 24/2[2(5)+(24-1)2]
=> 12(10+23×2)
=> 12(10+46)
=> 12(56)
=> 672.
That's the answer!
Hope it helps :)
Please mark me the brainliest.
Since it is given that the nth teem is given by an = 3+2n,
a1 = 3+2(1) => 3+2 => 5
a2 = 3+2(2) => 3+4 => 7
a3 = 3+2(3) => 3+6 =>9
Hence, a = 5; d = 7-5 => 2; n = 24
Now, we know that,
Sn = n/2[2a+(n-1)d]
So,
S24 = 24/2[2(5)+(24-1)2]
=> 12(10+23×2)
=> 12(10+46)
=> 12(56)
=> 672.
That's the answer!
Hope it helps :)
Please mark me the brainliest.
Lekahdek:
brainliest pls
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