Find the sum of 8 + 88 + 888 + … to n terms.

We have, 8 + 88 + 888 + … to n terms. = 8 ( 1 + 11 + 111 + … to n terms ) = 8 9 ( 9 + 99 + 999 + … to n terms ) = 8 9 ( ( 10 − 1 ) + ( 100 − 1 ) + ( 1000 − 1 ) + … to n terms ) = 8 9 ( ( 10 + 100 + 1000 + … to n terms ) − ( 1 + 1 + 1 + … to n terms ) = 8 9 ( ( 10 + 100 + 1000 + … to n terms ) − n ) We know that S n = a r n − 1 r − 1 , r > 1 where a = 10 is first term and common ratio r = 10 ∴ S n = 8 9 10 × 10 n − 1 10 − 1 - n ⇒ S n = 8 9 10 × 10 n − 1 9 - n ⇒ S n = 8 9 10 n + 1 − 10 9 - n ⇒ S n = 8 81 10 n + 1 − 10 − 9 n

Find the sum of 8 + 88 + 888 + … to n terms.

We have, 8 + 88 + 888 + … to n terms. = 8 ( 1 + 11 + 111 + … to n terms ) = 8 9 ( 9 + 99 + 999 + … to n terms ) = 8 9 ( ( 10 − 1 ) + ( 100 − 1 ) + ( 1000 − 1 ) + … to n terms ) = 8 9 ( ( 10 + 100 + 1000 + … to n terms ) − ( 1 + 1 + 1 + … to n terms ) = 8 9 ( ( 10 + 100 + 1000 + … to n terms ) − n ) We know that S n = a r n − 1 r − 1 , r > 1 where a = 10 is first term and common ratio r = 10 ∴ S n = 8 9 10 × 10 n − 1 10 − 1 - n ⇒ S n = 8 9 10 × 10 n − 1 9 - n ⇒ S n = 8 9 10 n + 1 − 10 9 - n ⇒ S n = 8 81 10 n + 1 − 10 − 9 n

Find the sum of the series 3 + 33 + 333 + … to n terms.

We have, 3 + 33 + 333 + … to n terms. = 3 ( 1 + 11 + 111 + … to n terms ) = 3 9 ( 9 + 99 + 999 + … to n terms ) [multiply and divide the expression by 9 ] = 3 9 ( ( 10 − 1 ) + ( 100 − 1 ) + ( 1000 − 1 ) + … to n terms ) = 3 9 ( ( 10 + 100 + 1000 + … to n terms ) − ( 1 + 1 + 1 + … to n terms ) ) = 3 9 ( ( 10 + 100 + 1000 + to n terms ) − n ) We know that S n = a r n − 1 r − 1 , r > 1 where a = 10 is first term and r = 10 is common ratio. S n = 3 9 10 × 10 n − 1 10 − 1 - n = 3 9 10 n + 1 − 10 - 9 n 9 = 1 27 10 n + 1 − 9 n − 10

Find the sum of the series 0.7 + 0.77 + 0.777 + … to n terms.

We have, 0.7 + 0.77 + 0.777 + … to n terms. = 7 ( 0.1 + 0.11 + 0.111 + … to n terms ) = 7 9 ( 0.9 + 0.99 + 0.999 + … to n terms ) [multiply and divide by 9 ] = 7 9 ( ( 1 − 0.1 ) + ( 1 − 0.01 ) + ( 1 − 0.001 ) + … to n terms ) = 7 9 ( ( 1 + 1 + 1 + … to n terms ) − ( 0.1 + 0.01 + 0.001 + … to n terms ) ) = 7 9 ( n − ( 0.1 + 0.01 + 0.001 + … to n terms ) ) We know that S n = a 1 − r n 1 − r , r < 1 where a = 0 . 1 is first term and r = 0 . 1 is common ratio S n = 7 9 n - 0.1 × 1 − 0.1 n 1 − 0.1 = 7 9 n - 1 − 0.1 n 9

E M B I B E

WBCHSE Mathematics for Class 11>Chapter 9 - Sequence and Series>EXERCISE 7G>Q 12

HARD

11th West Bengal Board

IMPORTANT

Earn 100

Find the sum of $8 + 88 + 888 + \dots to n terms.$

Important Questions on Sequence and Series

HARD

11th West Bengal Board

IMPORTANT

WBCHSE Mathematics for Class 11>Chapter 9 - Sequence and Series>EXERCISE 7G>Q 13

Find the sum of the series

3 + 33 + 333 + \dots to n terms.

HARD

11th West Bengal Board

IMPORTANT

WBCHSE Mathematics for Class 11>Chapter 9 - Sequence and Series>EXERCISE 7G>Q 14

Find the sum of the series

0.7 + 0.77 + 0.777 + \dots to n terms.

MEDIUM

11th West Bengal Board

IMPORTANT

WBCHSE Mathematics for Class 11>Chapter 9 - Sequence and Series>EXERCISE 7G>Q 15

Evaluate

\sum_{k = 1}^{11} (2 + 3^{k})

.

MEDIUM

11th West Bengal Board

IMPORTANT

WBCHSE Mathematics for Class 11>Chapter 9 - Sequence and Series>EXERCISE 7G>Q 16

The inventor of the chessboard suggested a reward of one grain of wheat for the first square; two grains for second; four grains for third, and so on, doubling the number of grains for subsequent squares. How many grains would have to be given to the inventor? (Note that there are

64

squares in the chessboard.)

MEDIUM

11th West Bengal Board

IMPORTANT

WBCHSE Mathematics for Class 11>Chapter 9 - Sequence and Series>EXERCISE 7G>Q 17

A man writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they move the chain similarly. Assuming that the chain is not broken and it costs