Math, asked by vijaya3484, 1 year ago

plz solve the questions​

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Answered by MrBasic
0

Given,

a_{n}=\frac{2^{n-1}}{5^{n-2}}\\=\frac{2^{n-1}}{5^{n-1}\times5^{-1}}}\\=5(\frac{2}{5})^{n-1}

Which is of the form ar^{n-1}, where a=5, r=\frac{2}{3}

Indeed this sequence is a G.P

Now,

Using the G.P summation to kth terms formula,

S_k=\frac{a(r^k-1)}{r-1}\\\\=5\frac{(\frac{2}{5})^k-1}{\frac{2}{5}-1}\\\\=5\frac{(\frac{2}{5})^k-1}{\frac{2-5}{5}}\\\\=\frac{25}{3}[1-(\frac{2}{5})^k]

Then,

S_\infty=\lim_{k\to\infty}\frac{25}{3}[1-(\frac{2}{5})^k]\\\\=\frac{25}{3}\lim_{k\to\infty}[1-(\frac{2}{5})^k]\\\\=\frac{25}{3}[1-\lim_{k\to\infty}(\frac{2}{5})^k]\\\\=\frac{25}{3}[1-0]\\\\\left[(\frac{2}{5})^k\to\:0\:as\:k\to\infty\right]\\\\=\frac{25}{3}\\\\\therefore S_\infty=\frac{25}{3}

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