Math, asked by rainyroy362, 19 hours ago

Consider the arithmetic sequence 100, 109, 118 · · ·
a) What is the common difference ?
b) What is the remainder on dividing the terms of the arithmetic sequence by 9
c) Which is the first four digit term of this sequence ?
d) Find the sum of all three digit terms of this sequence

Answers

Answered by pulakmath007
1

SOLUTION

GIVEN

Consider the arithmetic sequence 100, 109, 118 · · ·

TO DETERMINE

a) The common difference

b) The remainder on dividing the terms of the arithmetic sequence by 9

c) The first four digit term of this sequence

d) The sum of all three digit terms of this sequence

EVALUATION

Here the given arithmetic sequence is

100, 109, 118 · · ·

First term = a = 100

Common Difference = d = 109 - 100 = 9

a) The common difference = 9

b) The nth term of the sequence

= a + (n - 1)d

= 100 + 9( n - 1 )

= 100 + 9n - 9

= 9n + 91

= 9( n + 10 ) + 1

Now 9( n + 10 ) is divisible by 9

Hence the required Remainder = 1

c) The smallest 4 digit number = 1000

By the given condition

9n + 91 ≥ 1000

⇒ 9n ≥ 909

⇒ n ≥ 101

The first four digit term of this sequence

= 101th term of the sequence

= 1000

d) The sum of all three digit terms of this sequence

= 100 + 109 + 118 + . . . + 100th term

= 100 + 109 + 118 + . . . + 991

\displaystyle \sf{ = \frac{100}{2} (100 + 991) }

\displaystyle \sf{ = 50 \times 1091}

= 54550

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Answered by HarshitJaiswal2534
1

Step-by-step explanation:

SOLUTION

GIVEN

Consider the arithmetic sequence 100, 109, 118 · · ·

TO DETERMINE

a) The common difference

b) The remainder on dividing the terms of the arithmetic sequence by 9

c) The first four digit term of this sequence

d) The sum of all three digit terms of this sequence

EVALUATION

Here the given arithmetic sequence is

100, 109, 118 · · ·

First term = a = 100

Common Difference = d = 109 - 100 = 9

a) The common difference = 9

b) The nth term of the sequence

= a + (n - 1)d

= 100 + 9( n - 1 )

= 100 + 9n - 9

= 9n + 91

= 9( n + 10 ) + 1

Now 9( n + 10 ) is divisible by 9

Hence the required Remainder = 1

c) The smallest 4 digit number = 1000

By the given condition

9n + 91 ≥ 1000

⇒ 9n ≥ 909

⇒ n ≥ 101

The first four digit term of this sequence

= 101th term of the sequence

= 1000

d) The sum of all three digit terms of this sequence

= 100 + 109 + 118 + . . . + 100th term

= 100 + 109 + 118 + . . . + 991

\displaystyle \sf{ = \frac{100}{2} (100 + 991) }

\displaystyle \sf{ = 50 \times 1091}

= 54550

━━━━━━━━━━━━━━━━

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