Question

3, 33, 333, 3333, ......... या क्रमाने 9 संख्या घेऊन बेरीज केल्यास दशकस्तानी कोनता अंक येईल?​

Answer 1

Answer:

6 will  be the digit at Ten's Place

Step-by-step explanation:

3 + 33 + 333  +  3333+ ..................................................+ 333333333

We need to find digit at Ten's Place

so just add unit's & Ten' Digit

there will be 3 , Nine times at unit's places

& there will be 3 , Eight times at Ten's place  

3 , Nine times at unit's places = 3  * 9 = 27

7 at unit place & two is carried forward for Ten's Place

Tens Place = 2 +  3*8  = 26

6 will  be the digit at Ten's Place

Sum would be 370,370,367

Answer 2

Answer:

1111111083

Step-by-step explanation:

Since we have to add 3 nine terms.

The  series given to us is 3, 33, 333, 3333, .........

We have to calculate the value till 9 terms, as mentioned in the question.

Now, we will be adding all the terms by taking 3 common.

The series would look like as:

3 (1 + 11 + 111 + 1111 + 11111 +.............+ 111111111)

We will subtract each term from 10;

$3 \times [(10-1) + (10^{2} -1)+(10^{3} -1)+..........+(10^{9} -1)]$

$3 \times [10 + 10^{2} + 10^{3} + 10^{4} +...........+ 10^{9} ]$ - 27

Since the above term are in gp;

So, we will be using the sum for gp;

We will be getting the sum as $\frac{10 \times (10^{2} -1)}{9} - 27$

=$\frac{10 \times 999999999}{9} -27$

$\frac{9999999990}{9} - 27$

1111111110 - 27

1111111083

So, the answer will be 1111111083