Question

Find the sum of series :7+77+777+.........to n term.​

Answer 1

$\huge{\underline{\underline{\red{\bf{Given:}}}}}$

• A series is given 7+77+777+777..........n terms

$\rule{200}4$

$\huge{\underline{\underline{\red{\bf{To\: Find:}}}}}$

• The sum of the series.

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$\huge{\underline{\underline{\red{\bf{Answer:}}}}}$

Given series is 7+77+777+7777+......n terms.

Let the sum of series be demoted by S.

$\sf{\implies S = 7+77+777+7777+......n .}$

$\sf{\implies S = 7 ( 1 + 11 + 111 + 1111 + .......n ).}$

$\sf{\implies S = \dfrac{7}{9} \times 9( 1 + 11 + 111 + 1111 + ... n ).}$

$\sf{\implies S = \dfrac{7}{9} \times( 9 + 99 + 999 + 9999 + ...n ).}$

$\sf{\implies S =\dfrac{7}{9} \times [(10-1) + (100-1)+(1000-1)+......n ]}$

$\sf{\implies S = 7/9[10+100+1000+....+n -1-1-1...-n]}$

Now this 10+100+1000 .....n are in GP,

And sum of n terms of GP is given by ,

$\large{\underline{\boxed{\red{\bf{\leadsto S_{n}=\dfrac{ a(r^{n}-1)}{r-1}}}}}}$

where ,

• a is first term.
• r is common ratio.
• n is number of terms.

Using this formula , we have ;

⇒S = 7/9 [ $\sf{\dfrac{10(10^n-1)}{9}}$ - n].

$\purple{\bf{ \longmapsto S =\dfrac{70(10^n -1)}{81}-\dfrac{7n}{9}}}$