Question

the sum of the Arithmetic progression 34,32,30, ....,10 is

Answer 1

Answer:

286

Step-by-step explanation:

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Answer 2

Answer:

$The \:sum \:of \:the\\ Arithmetic \:progression\\ 34,32,30, ....,10 \:is\:286$

Step-by-step explanation:

$Given \:A.P : 34,32,30,...,10$

$First\:term (a) = 34\\common \: difference (d) = a_{2}-a_{1}\\=32-34\\=-2$

$last \:term (a_{n})=10$

$\implies a+(n-1)d = 10$

$\implies 34+(n-1)(-2)=10$

$\implies (n-1)(-2)=10-34$

$\implies (n-1)(-2)=-24$

$\implies n-1=\frac{-24}{-2}$

$\implies n-1=12$

$\implies n = 12+1 = 13$

$Now,\\</p><p>Sum \:of \: n \: terms (S_{n})\\=\frac{n}{2}[2a+(n-1)d]\\=\frac{13}{2}[2\times 34+(13-1)(-2)]\\=\frac{13}{2}[68-24]\\=\frac{13}{2}\times 44\\=13\times 22\\= 286$

Therefore,

$The \:sum \:of \:the\\ Arithmetic \:progression\\ 34,32,30, ....,10 \:is\:286$

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