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Sequence and Series

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Question : 19 of 86

Marks: +1, -0

Find the sum of first twenty five terms of AP series whose nth term is

\left( \; \frac{n}{5} + 2 \right)

105
115
125
135

Solution:

Concept:

The sum of the first

n

terms of an AP can be found using

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

, where

a

is the first term and

d

is the common difference.

Explanation:

Given

T_n = \frac{n}{5} + 2

.

First term

a = T_1 = \frac{1}{5} + 2 = \frac{11}{5}

.

Second term

T_2 = \frac{2}{5} + 2 = \frac{12}{5}

.

Common difference

d = T_2 - T_1 = \frac{12}{5} - \frac{11}{5} = \frac{1}{5}

.

Number of terms

n = 25

.

Sum

S_{25} = \frac{25}{2}\left[2 \cdot \frac{11}{5} + (25-1)\cdot \frac{1}{5}\right] = \frac{25}{2}\left[\frac{22}{5} + \frac{24}{5}\right] = \frac{25}{2} \cdot \frac{46}{5} = \frac{25 \times 46}{10} = \frac{1150}{10} = 115

.

Answer:

115

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