Mathematics, 06.07.2021 14:00 bakerx8529
The sum of the first ten terms of an arithmetic progression consisting of
positive integer terms is equal to the sum of the 20th, 21st and 22nd term.
If the first term is less than 20, find how many terms are required to give
a sum of 960.
Answers: 3
![\Rightarrow \dfrac{10}{2}[2a+(10-1)d]=[a+19d]+[a+20d]+[a+21d]\\\\\Rightarrow 5[2a+9d]=3a+60d\\\Rightarrow 10a+45d=3a+60d\\\Rightarrow 7a=15d](/tpl/images/1389/6372/beeaf.png)
![\Rightarrow 960=\dfrac{n}{2}[2a+(n-1)d]\\\\\Rightarrow 1920=n[2a+(n-1)\cdot \dfrac{7}{15}a]\\\\\Rightarrow 28,800=n[30a+7a(n-1)]\\\\\Rightarrow a=\dfrac{28,800}{n[30+7n-7]}\\\\\Rightarrow a=\dfrac{28,800}{n[23+7n]}](/tpl/images/1389/6372/5f5c2.png)
![\therefore \dfrac{28,800}{n[23+7n]}](/tpl/images/1389/6372/18e89.png)
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The sum of the first ten terms of an arithmetic progression consisting of
positive integer terms is...
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