Question

find the sum of first n terms of 4+44+444+4444+....... in geometric progression it is urgent explain it well for class 10

Answer 1

Its easy to see that this is a diverging sequence and if you keep following it up to infinity as you show by the dots following the 4th term, the answer will approach infinity. However if you want to sum it up to n terms, there is an easy way provided you know the sum of a Geometric Progression.

Rewrite 4 + 44 + 444 + 4444 +…..up to n terms as

4/9 ( 9 + 99 + 999 + 9999 +….up to n terms)

4/9 ( 10 - 1 + 100 - 1 + 1000 - 1 + 10000 - 1 +…..up to n terms)

4/9 ( 10 + 100 + 1000 + 10000 +…up to n terms - n)

The next bit is easy, just apply the formula for calculating sum of a G.P. which is a(r^n - 1)/(r - 1) where a is the first term, r the common ratio and n the number of terms in the sequence.

So you finally get 4/9[ 10(10^n - 1)/9] or 40/81[ 10^n - 1 ]

Answer 2

S= 4+44+444+........n terms   =4(1+11+111+.........n terms)   = 49(9+99+999+........n terms)   =49[(10−1)+(102−1)+(103−1)+.........(10n−1)]   =49[(10+102+103+........10n)−(1+1+1......n times)]   =49{10[10n−110−1]−n(1)}    =49[(109)(10n−1)−n]    = 481[10n+1−10−9n]