Sum of first n terms is 3n2 + n. Find the 22nd term.

First Term of the AP (a) = 5 Common difference (d) = 8 - 5 = 3 Last term = a40 = a + (40 - 1) d = 5 + 39 × 3...

Let the numbers be: a - 3d, a - d, a + d, a + 3d a - 3d + a - d + a + d + a + 3d...

Sn = 3n2 + n  S22 = 3(22)2 + 22 S22 = 1452 + 22 ∴ S22 = 1474 S21 = 3(21)2 + 21 S21 = 1344 a22 = S22 - S21  ∴ a22 = 1474 - 1344 ∴ a22 = 130 Hence, the 22nd term of A.P is 130

Correct Answer - 1,-1,2

1. (c)  2. (e)  3. (a)  4. (b)  5. (d)

1. 15  2. 16

Yes, it is valid although the registrar issued the certification on 24th but on the certificate 22nd August is written. So in the eyes of law, the company was registered...

Solution: As the 7th term exceeds the 5th term by 12, so the 5th term will exceed the 3rd term by 12 as well So, n3 = 16 n5 = 28 n7 = 40 n4 or n6 can be...