Question

Find the sum to n terms of the sequence ⁸'⁸⁸'⁸⁸⁸'⁸⁸⁸⁸...​

Answer 1

★ Concept :-

Here the concept of Geometric Progression (GP) has been used. We see that we are given a sequence and we have to find the sum of n terms. Firstly we have to find out the common ratio of the sequence. By formula we can find it . And then we have to compare that common ratio and apply the formula of sum of n terms and thus find the answer.

Let's do it !!

______________________________________

★ Formula Used :-

$\;\boxed{\sf{\pink{Common\;Ratio,\;r\;=\;\bf{\dfrac{a_{2}}{a_{1}}\;=\;\dfrac{a_{3}}{a_{2}}}}}}$

$\;\boxed{\sf{\pink{S_{n}\;=\;\bf{\dfrac{a(r^{n}\;-\;1)}{r\;-\;1}}}}}$

______________________________________

★ Solution :-

Given,

Series : 8, 88, 888, 8888, ... upto n terms

→ Series : 8(1, 11, 111, 1111, ... upto n terms)

→ Series : 8(9/9, 99/9, 999/9, 9999/9, ...)

→ Series : 8[(10-1)/9, (100-1)/9, (1000-1)/9, (10000-1)/9 ... upto n terms ]

→ Series : 8/9[(10-1), (100-1), (1000-1), (10000-1), ... upto n terms]

→ Series : 8/9[(10-1), (10²-1), (10³-1), (10⁴-1), ... upto n terms]

Since we have to find sum, so adding all terms, we get

→ Series : 8/9[(10-1) + (10²-1) + (10³-1) + (10⁴-1) +...+ upto n terms]

Now seperating 1 from main terms, we get

→ Series : 8/9[(10 + 10² + 10³ + 10⁴ +...+ upto n terms) - (1 + 1 + 1 +...+ upto n terms)]

→ Series : 8/9[(10 + 10² + 10³ + 10⁴ +...+ upto n terms) - n(1)]

Clearly we see that terms in sequence that is series of 10 is related to each other by a factor and thus it is a GP. So, this can be written as :-

» G.P. : 10, 10², 10³, 10⁴,..., n terms

Here,

» First term = a₁ = 10

» Second term = a₂ = 10² = 100

» Third term = a₃ = 10³ = 1000

---------------------------------------------

~ For common ratio of this G.P ::

We know that,

$\;\sf{\rightarrow\;\;Common\;Ratio,\;r\;=\;\bf{\dfrac{a_{2}}{a_{1}}\;=\;\dfrac{a_{3}}{a_{2}}}}$

By applying values, we get

$\;\sf{\rightarrow\;\;Common\;Ratio,\;r\;=\;\bf{\dfrac{100}{10}\;=\;\dfrac{1000}{100}}}$

$\;\sf{\rightarrow\;\;Common\;Ratio,\;r\;=\;\bf{10\;=\;10}}$

$\;\bf{\rightarrow\;\;Common\;Ratio,\;r\;=\;\bf{\red{10}}}$

----------------------------------------------

~ For sum of this G.P. ::

We know that,

$\;\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{a(r^{n}\;-\;1)}{r\;-\;1}}}$

By applying values, we get

$\;\sf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{10(10^{n}\;-\;1)}{10\;-\;1}}}$

$\;\bf{\rightarrow\;\;S_{n}\;=\;\bf{\orange{\dfrac{10(10^{n}\;-\;1)}{9}}}}$

----------------------------------------------

~ For the sum of the given sequence ::

By applying this, we get

$\;\bf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{8}{9}\bigg[\dfrac{10(10^{n}\;-\;1)}{9}\;-\;n\bigg]}}$

$\;\bf{\rightarrow\;\;S_{n}\;=\;\bf{\dfrac{8}{9}\bigg[\dfrac{10(10^{n}\;-\;1)\;-\;9n}{9}\bigg]}}$

$\;\bf{\rightarrow\;\;\blue{S_{n}\;=\;\bf{\dfrac{8}{9}\bigg[\dfrac{10(10^{n}\;-\;1)\;-\;9n}{9}\bigg]}}}$

This is the required answer.

$\;\underline{\boxed{\tt{Required\;\:Sum\;=\;\bf{\purple{\dfrac{8}{9}\bigg[\dfrac{10(10^{n}\;-\;1)\;-\;9n}{9}\bigg]}}}}}$

________________________________

★ More to know :-

$\;\tt{\leadsto\;\;Geometric\;Mean\;=\;\sqrt{a_{1}\:a_{2}}}$

$\;\tt{\leadsto\;\;N^{th}\;term,\;a_{n}\;=\;ar^{n\:-\:1}}$

• Formulas for A.P. ::

$\;\tt{\leadsto\;\;Common\; Difference\;=\;a_{2}\:-\:a_{1}}$

$\;\tt{\leadsto\;\;N^{th}\;term,\;a_{n}\;=\;a\:+\:(n\:-\:1)d}$

$\;\tt{\leadsto\;\;Arithmetic\;Mean\;=\;\dfrac{a_{1}\:+\:a_{2}}{2}}$

$\;\tt{\leadsto\;\;S_{n}\;=\;\dfrac{n}{2}\:[a\:+\:a\:+\:(n\:-\:1)d]}$

________________________________