Test Index

Sequence and Series

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Question : 56 of 86

Marks: +1, -0

The

p^{\text{th}}

AP

\frac{3p-1}{6},

the sum of the first

n

A P

is

$n(3n+1)$
$\frac{n}{12}(3n+1)$
$\frac{n}{12}(3n-1)$
None of these

Solution:

Concept:

The

p

th term of an AP is linear in

p

:

T_p = dp + (a-d)

.

Explanation:

Given

T_p = \frac{3p-1}{6} = \frac{1}{2}p - \frac{1}{6}

.

Comparing with

T_p = dp + (a-d)

gives

d = \frac{1}{2}

and

a-d = -\frac{1}{6}

.

Thus

a = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}

.

Sum of first

n

terms:

S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}\left[2\cdot\frac{1}{3} + (n-1)\cdot\frac{1}{2}\right]

.

Simplify:

S_n = \frac{n}{2}\left[\frac{2}{3} + \frac{n-1}{2}\right] = \frac{n}{2}\cdot\frac{4 + 3(n-1)}{6} = \frac{n}{2}\cdot\frac{3n+1}{6} = \frac{n(3n+1)}{12}

.

Answer:

\frac{n}{12}(3n+1)

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