Find the sum to 20 terms of the A.P 6,10,14, …..
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Answered by
1
Answer:
a=2
d=6-2=4
n=20
s=n/2[2a+(n-1).d]
s=20/2[2.2+(20-1).4]
s=10[4+(19).4]
s=10[4+76]
s=10[80]
s=10*80
s=800
a = 2
d = 6 - 2 = 4
Sn = n/2 [ 2a + (n-1) d]
= 20/2 [ 2(2) + 19(4)]
= 10 ( 4 + 76)
= 10 × 80
= 800
Step-by-step explanation:
Answered by
0
Very simple
As first term of AP is 6 & common difference is 4
We can write general formula for nth term as
Tn = 6 + 4*(n - 1) = 4*n + 2
So 6th term is 4*6 + 2 = 26 &
12th term is 4*12 + 2 = 50
So using summation formula for AP
Sn = n/2 ( first term + last term )
Sum of 7 terms from 6th to twelth is
S7 = 7/2 ( 26 + 50 ) = 7 × 38 = 266
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As first term of AP is 6 & common difference is 4
We can write general formula for nth term as
Tn = 6 + 4*(n - 1) = 4*n + 2
So 6th term is 4*6 + 2 = 26 &
12th term is 4*12 + 2 = 50
So using summation formula for AP
Sn = n/2 ( first term + last term )
Sum of 7 terms from 6th to twelth is
S7 = 7/2 ( 26 + 50 ) = 7 × 38 = 266
Mark me as brainliest
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