find the sum of the first 7 terms of AP sequence 27,30,33,
Answers
=63
2
7
[2a+(7−1)d]=63
14a+42d=126
a+3d=9 --- (i)
Sum of next seven term is 161
S
14
−S
7
=161
2
14
[2a+13d]−63=161
7(2a+13d)=224
a+
2
13
d=16 ---- (ii)
on subtracting (i) from (ii)
we get,
d=2
from (i)
a=3
a
25
=a+24d
=3+24×2=51
Answer:
Sum of first 7 terms of given A.P = 252.
Step-by-step explanation:
Arithmetic Progression:
- A series of numbers is called a "arithmetic progression" (AP) when any two subsequent numbers have a constant difference.
- Arithmetic Sequence is another name for A.P.
- A common difference between two succeeding words (let's say 1 and 2) in the natural number sequence 1, 2, 3, 4, 5, 6,..., for instance, is equal to 1.
- The common difference between two consecutive words will always equal 1, even in the situation of odd and even numbers.
Given terms of AP are 27, 30, 33
First term = a₁ = 27
Second term = a₂ = 30
Third term = a₃ = 33
Difference between the terms is the common difference.
Common difference d = 30-27 = 3
Here a = a₁ = 27, a₂ = 30, a₃ = 33 and d = 3
Sum of first n terms = Sₙ = (n/2)[2a+(n-1)d]
So, Sum of first 7 terms is
S₇ = (7/2)[2(27)+(7-1)3]
= (3.5)[54+6(3)]
= (3.5)[54+18]
= (3.5)(72)
S₇ = 252
Hence, Sum of first 7 terms of given A.P = 252.
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