Answer :
The 28th term of a given arithmetic sequence is 57.
Consider that "a" is the first term and "d" is the common difference.
The sum of the first n terms can be calculated as:
Sₙ = n/2(2a+(n−1)d)
Given that the sum of the first 7 terms of an A.P. is 63.
And the sum of the next 7 terms is 161.
Sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms
= 63 + 161 = 224
Now,
S₇ = [2a + (7 − 1)d]
63 = (2a + 6d)
18 = 2a + 6d
2a + 6d = 18 ....(1)
Also,
S₁₄ = [2a + (14 − 1)d]
224 = 7(2a + 13d)
32 = 2a + 13d
2a + 13d = 32 ....(2)
Subtract (1) from (2) to get:
13d − 6d = 32 − 18
7d = 14
d = 2
Substitute the value of d = 2 into equation (1),
2a = 18 − 6 × 2
2a = 18 − 12
2a = 6
a = 3
Also, nth term can be expressed as:
an = a + (n − 1)d
a₂₈ = 3 + (28 − 1)2
= 3 + 27 × 2
= 57
Thus, the 28th term of a given arithmetic sequence is 57.
Learn more about arithmetic sequence here:
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The complete question is as follows:
The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.