# How many terms from the progression 3, 5, 7, 9, ... must be taken in order that their sum will be 2600?

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**Problem**

How many terms from the progression 3, 5, 7, 9, ... must be taken in order that their sum will be 2600?

A. 80 | C. 50 |

B. 60 | D. 70 |

**Answer Key**

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[ C ]

**Solution**

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3, 5, 7, 9, ... ← is an AP with

$S = \dfrac{n}{2}\left[ 2a_1 + (n - 1)d \right]$

*d*= 2$S = \dfrac{n}{2}\left[ 2a_1 + (n - 1)d \right]$

$2600 = \dfrac{n}{2}\left[ 2(3) + (n - 1)(2) \right]$

$2600 = 3n + n(n - 1)$

$n^2 + 2n - 2600 = 0$

$(n - 50)(n + 52) = 0$

$n = 50 ~ \text{and} ~ -52$

Use $n = 50$ ← *answer*

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