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47

Expert-verifiedFound in: Page 816

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

Find each sum

$4+4.5+5+5.5+...+100$

The solution is ${S}_{193}=10036$

$4+4.5+5+5.5+...+100$

We have,$4+4.5+5+5.5+...+100$

So, ${a}_{1}=4,d=0.5,{a}_{n}=100$

Now use formula,

${a}_{n}={a}_{1}+(n-1)d\phantom{\rule{0ex}{0ex}}\Rightarrow 100=4+(n-1)0.5\phantom{\rule{0ex}{0ex}}\Rightarrow 100=4+0.5n-0.5\phantom{\rule{0ex}{0ex}}\Rightarrow 0.5n=96.5\phantom{\rule{0ex}{0ex}}\Rightarrow n=193$

To find sum ${S}_{193}$

${S}_{193}=\frac{193}{2}[4+100]\phantom{\rule{0ex}{0ex}}=10036$

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