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Progression Practice Test 1

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Question : 17 of 23

Marks: +1, -0

Find the sum of 125 terms of the AP

\frac{1}{6},\frac{2}{7},\frac{3}{7},\ldots

1,125
$\frac{1025}{7}$
1,025
$\frac{1125}{7}$
1,126

Solution: 👈: Video Solution

Concept:

The sum of the first

n

terms of an AP is

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

, where

a

is the first term and

d

is the common difference.

Explanation:

The given AP is

\frac{1}{7},\frac{2}{7},\frac{3}{7},\ldots

First term

a=\frac{1}{7}

.

Common difference

d=\frac{2}{7}-\frac{1}{7}=\frac{1}{7}

.

Number of terms

n=125

.

Substitute in the sum formula:

S_{125}=\frac{125}{2}\left[2\cdot\frac{1}{7}+(125-1)\cdot\frac{1}{7}\right]

S_{125}=\frac{125}{2}\left[\frac{2}{7}+\frac{124}{7}\right]

S_{125}=\frac{125}{2}\cdot\frac{126}{7}

S_{125}=\frac{125}{2}\cdot 18

S_{125}=125\cdot 9=1125

Answer:

1125

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