ARITHMETIC PROGRESSION 1

The 1st term of an A.P. is 90 and the common difference is 33. Find the 10th term.

Solution

A1= 90, d = 33

An = A1 + (n-1)d

A10 = A1 + (10-1)d

A10 = A1 + 9d

A10 = 90 + (9x33) [ after substituting A1= 90, d = 33 as given above]

∴ A10 = 90 + 297

A10 = 387

Hence the 10th term is 387.


TRY THIS……………

The 1st term of an A.P. is 170 and the common difference is 39. Find the 6th term.