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Question: p<sup>th</sup> term of the series \(\left( 3 - \frac { 1 } { n } \right) + \left( 3 - \frac { 2 } {...

pth term of the series (3−1n)+(3−2n)+(3−3n)+…\left( 3 - \frac { 1 } { n } \right) + \left( 3 - \frac { 2 } { n } \right) + \left( 3 - \frac { 3 } { n } \right) + \ldots will be.

A

(3+pn)\left( 3 + \frac { p } { n } \right)

B

(3−pn)\left( 3 - \frac { p } { n } \right)

C

(3+np)\left( 3 + \frac { n } { p } \right)

D

(3−np)\left( 3 - \frac { n } { p } \right)

Answer

(3−pn)\left( 3 - \frac { p } { n } \right)

Explanation

Solution

Given series (3−1n)+(3−2n)+(3−3n)+………\left( 3 - \frac { 1 } { n } \right) + \left( 3 - \frac { 2 } { n } \right) + \left( 3 - \frac { 3 } { n } \right) + \ldots \ldots \ldots (A.P.)

Therefore common difference

d=(3−2n)−(3−1n)=−1nd = \left( 3 - \frac { 2 } { n } \right) - \left( 3 - \frac { 1 } { n } \right) = - \frac { 1 } { n } and first term a=(3−1n)a = \left( 3 - \frac { 1 } { n } \right)

Now pthp ^ { t h } term of the series =a+(p−1)d= a + ( p - 1 ) d

=(3−1n)+(p−1)(−1n)=3−1n+1n−pn=(3−pn)= \left( 3 - \frac { 1 } { n } \right) + ( p - 1 ) \left( - \frac { 1 } { n } \right) = 3 - \frac { 1 } { n } + \frac { 1 } { n } - \frac { p } { n } = \left( 3 - \frac { p } { n } \right) .

Trick : This question can also be done by inspection first −1n- \frac { 1 } { n }, second −2n- \frac { 2 } { n }, third −3n- \frac { 3 } { n } , therefore, pthp ^ { t h } will be −pn- \frac { p } { n }. Hence the result (3 is constant).