Question

p^th term of the series $(3-\frac{1}{n})+(3-\frac{2}{n})+(3-\frac{3}{n})+....$ will be

• A. $(3+\frac{P}{n})$
• B. $(3-\frac{P}{n})$
• C. $(3+\frac{n}{P})$
• D. $(3-\frac{n}{P})$

Answer 1

The correct option is B $(3-\frac{P}{n})$

Given series $(3-\frac{1}{n})+(3-\frac{2}{n})+(3-\frac{3}{n})+....$ (A.P.)

Therefore common difference

d = $(3-\frac{2}{n})+(3-\frac{1}{n})=\frac{1}{n}$ and first term a = $(3-\frac{1}{n})$
Now p^th term of the series = a + (p - 1)d = $(3-\frac{1}{n})+(p-1)(-\frac{1}{n})=3-\frac{1}{n}+\frac{1}{n}-\frac{P}{n}=(3-\frac{P}{n}).$

Trick : This question can also be done by inspection i.e. first $-\frac{1}{n},$ second $-\frac{2}{n},$ third $-\frac{3}{n},$ therefore, pth will be $\frac{P}{n}.$ Hence the result (3 is constant).