The sum of first 7 terms of an A.P. is 63 and the sum of its next 7 te

1.	The sum of 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.
Answer» Let’s take a to be the first term and d to be the common difference. And we know that, sum of first n terms S_n = \(\frac{n}{2}\)(2a + (n − 1)d) Given that sum of the first 7 terms of an A.P. is 63. S₇ = 63 And sum of next 7 terms is 161. So, the sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms S₁₄= 63 + 161 = 224 Now, having S₇ = \(\frac{7}{2}\)(2a + (7 − 1)d) ⟹ 63(2) = 7(2a + 6d) ⟹ 9 × 2 = 2a + 6d ⟹ 2a + 6d = 18 . . . . (1) And, S₁₄ = \(\frac{14}{2}\)(2a + (14 − 1)d) ⟹ 224 = 7(2a + 13d) ⟹ 32 = 2a + 13d …. (2) Now, subtracting (1) from (2), we get ⟹ 13d – 6d = 32 – 18 ⟹ 7d = 14 ⟹ d = 2 Using d in (1), we have 2a + 6(2) = 18 2a = 18 – 12 a = 3 Thus, from n^th term ⟹ a₂₈= a + (28 – 1)d = 3 + 27 (2) = 3 + 54 = 57 Therefore, the 28^th term is 57.

The sum of 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.

Answer»

Let’s take a to be the first term and d to be the common difference.

And we know that, sum of first n terms

S_n = \(\frac{n}{2}\)(2a + (n − 1)d)

Given that sum of the first 7 terms of an A.P. is 63.

S₇ = 63

And sum of next 7 terms is 161.

So, the sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms

S₁₄= 63 + 161 = 224

Now, having

S₇ = \(\frac{7}{2}\)(2a + (7 − 1)d)

⟹ 63(2) = 7(2a + 6d)

⟹ 9 × 2 = 2a + 6d

⟹ 2a + 6d = 18 . . . . (1)

And,

S₁₄ = \(\frac{14}{2}\)(2a + (14 − 1)d)

⟹ 224 = 7(2a + 13d)

⟹ 32 = 2a + 13d …. (2)

Now, subtracting (1) from (2), we get

⟹ 13d – 6d = 32 – 18

⟹ 7d = 14

⟹ d = 2

Using d in (1), we have

2a + 6(2) = 18

2a = 18 – 12

a = 3

Thus, from n^th term

⟹ a₂₈= a + (28 – 1)d

= 3 + 27 (2)

= 3 + 54 = 57

Therefore, the 28^th term is 57.