Question

Challenge to Maths Masters

Find the sum:
$\: \: \: \boxed{ \boxed{ \bold{\orange{ S = \frac{5}{2} + \frac{55}{22} + \frac{555}{222} + .... \tt \: nth \: term}}}}$
Hint:
${S}_{GP} + {S}_{AP} =S$
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Answer 1

$\texttt{\textsf{\large{\underline{Topic}:}}}$

• Series.

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Given:

$\tt\implies S=\dfrac{5}{2}+\dfrac{55}{22}+\dfrac{555}{222}+ \cdot\cdot\cdot$

$\tt\implies S=\dfrac{5}{2}+\dfrac{11\times5}{11\times2}+\dfrac{111\times5}{111\times2}+ \cdot\cdot\cdot$

We can simplify it and write it as:

$\tt\implies S=\dfrac{5}{2}+\dfrac{5}{2}+\dfrac{5}{2}+ \cdot\cdot\cdot$

Now, we can consider this as AP as well as GP.

Considering it as AP,

$\tt\implies a=\dfrac{5}{2}$

$\tt\implies d=0$  (Yes, an AP can have common difference as 0)

Using formula,

$\tt\implies S_{n} = \dfrac{n}{2}\big(2a+(n-1)d\big)$

Substituting the value in the formula, we get,

$\tt\implies S_{n} = \dfrac{n}{2}\big(2\times {}^{5}/_{2}+(n-1)\times0\big)$

$\tt\implies S_{n} = \dfrac{n}{2}\times5$

$\tt\implies S_{n} = \dfrac{5n}{2}$

Which is our required answer.

$\rule{200}{2}$

If we consider this as GP,

$\tt\implies a=\dfrac{5}{2}$

$\tt\implies r=1$

Using formula,

$\tt\implies S_{n}=\dfrac{a(1-r^{n})}{1-r}$

But r ≠ 1.

In this case, we will consider another formula.

$\tt\implies S_{n}=a+a+a+... n\ times = na$

$\tt\implies S_{n}=n\times\dfrac{5}{2}$

$\tt\implies S_{n}=\dfrac{5n}{2}$   (Answer)

$\texttt{\textsf{\large{\underline{Formulae To Know}:}}}$

In AP:

1. nᵗʰ term = a + (n - 1)d

2. Sum of first n terms = n(a + l)/2

Where,

1. a = First term.
2. d = Common difference.
3. l = Last term.

In GP:

1. nth term = arⁿ⁻¹

2. Sum of first n terms = a(1 - rⁿ)/(1 - r)

Where,

1. a = First term.
2. r = Common ratio.

Answer 2

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HERE IS UR ANSWER

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