Class 10 Math All topics Arithmetic Progressions

Find the sum of the following arithmetic progressions:

$41,36,31,.....$ to $12$ terms.

$41,36,31,.....$ to $12$ terms.

Solution: _{}^{}_{}^{}_{}^{}_{}^{}_{}^{}_{}^{}_{}^{}_{}^{}_{}^{}_{}^{}

Given A.P is

$41,36,31,...…..$

number of terms of A.P is $n=12$

first term of this A.P is $a_{1}=41$

second term of this A.P is $a_{2}=36$

common difference

$d=a_{2}−a_{1}=36−41=−5$

we know that sum of n term of an A.P is given by

$S_{n}=2n [2a+(n−1)d)]$

$⟹S_{12}=212 [2×41+(12−1)×−5]$

$⟹S_{12}=6[82+(11)×−5)]$

$⟹S_{12}=6[82−55]$

$⟹S_{12}=6[27]$

$⟹S_{12}=162$

hence the sum of 12 terms of given A.P is $162.$

Similar topics

introduction to trigonometry

functions

some applications of trigonometry

quadratic equations

surface areas and volumes

introduction to trigonometry

functions

some applications of trigonometry

quadratic equations

surface areas and volumes

Related Questions

Related Questions