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The sum of the first 7 terms of an arithmetic progression (AP) is 63, and the sum of its next 7 terms is 161. Find the 28th term of this AP.

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.

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